ALL OF GRADE 10 MATH IN ONLY 1 HOUR!!! | jensenmath.ca

with 22 practice questions i'm going to

teach you the entire grade 10 math

course

in just one hour the ministry of

education says it should take 110 hours

to learn all of the material

let me see if i can save you some time

this video will be a great way to review

the material you have already learned or

to get introduced to the types of

questions taught at this grade level

here we go

[Music]

example one solve the following linear

system in three ways by graphing it

substitution

and elimination with all three methods

we should get the same answer

but we'll go through how you do each of

them now when solving a linear system

basically what that means is we have two

linear equations we want to figure out

what's the values of the variables that

make both equations true we can do that

algebraically with substitution or

elimination

or graphically by graphing both lines

and seeing where the point of

intersection

is so let's do graphing first in order

to graph both of these lines it's

easiest if we rearrange them

into y equals mx plus b form so i'm

going to isolate y in both of these

equations

let me start with the x minus y equals 5

equation

let me isolate y it'll be easiest if i

isolate it to the right

and now you can see it's in y equals mx

plus b form

where m is the slope and in this case

our m value

is 1 and b is our y

intercept and in this case our b value

is negative 5. so we could use that

slope and y intercept

to graph the line so we would plot the y

intercept of negative five

and then use the slope of one remember

slope is rise over runs you may want to

rewrite that whole number one as a

fraction one over one

so rise one run 1 plot a point and keep

plotting points until you fill the grid

and then we'll connect them with a

straight line and we'll go through the

same process for the other line

let me isolate y i'll just move the 3x

term to the right becomes negative 3x

so it's clear that for this one the

slope is negative 3

and the y-intercept is 3. plot the

y-intercept

and then use the slope and remember any

whole number is over 1 so you can think

of this negative 3 as negative 3

over 1. so rise negative three run one

that means down three right one

and keep plotting points until you fill

your grid

connect them with a straight line and

then just find where these lines

intersect and that's the solution to the

system

these lines intersect right at that

point right there the point two negative

three that's our point of intersection

so we can say the solution is you could

write it one of two ways you could write

it as

a point the point two negative three

or you could write what x and y are

equal to you could write

x equals 2 y equals negative 3.

either would be acceptable ways of

writing the final answer let's solve

that same system

using elimination and substitution

elimination first

so in doing elimination you want the

equations in the format x plus y

equals the constant which both equations

are already in that format

now our goal here is to eliminate a

variable and the way we do that

is by making the coefficients of either

the x's or the y's

have the same absolute value now in this

example the coefficients of the y's

already have the same absolute value the

coefficient

of y in the first equation is negative 1

and the coefficient of y in the second

equation

is positive one those have the same

absolute value but because they're

opposite signs

adding the equations will make them

eliminate if they were the same sign

subtracting them would make them

eliminate but in this case since they're

the same absolute value but opposite

signs

if we add these two equations together

the y's will eliminate

watch let's add the equations and when

we add the equations we make sure we

collect the like terms that's why we

have them lined up like this

x plus 3x is 4x negative y plus y

is 0 that they eliminate which is what

we want

and 5 plus 3 is 8. so we have 4x equals

8 divide both sides by 4

we get x equals 2. so we have the value

of x

that make both equations the same but

what's the value of y we now need to sub

x equals 2 into either of the original

equations

it doesn't matter which one you plug it

into because that's the value of x that

makes both equations have the same value

of y

i'll choose the first one it looks

easier so sub 2 in for x

2 minus y equals 5. i'll isolate y

move the negative y to the right move

the 5 to the left i've got 2 minus 5

equals y negative 3 equals

y so notice my solution x equals 2

y equals negative 3. that's the same

thing we got by solving it graphically

let's try it again but using

substitution now when doing substitution

i often like to line them up side by

side like this

i'll still number the equations equation

one and equation two

when doing substitution what you want to

do is isolate

one of the variables in either of the

equations doesn't matter

i think in equation one the variable x

like

looks like it would be easy to isolate

since it already has a coefficient of 1.

so i'll pick this x to be the variable

that i'm going to isolate

so i'll move this negative y to the

right and it becomes positive y so i

have

x equals 5 plus

y now what you do is you take

what the variable you isolated is equal

to so it says x equals 5 plus y

you take that and plug it into the other

equation for x

because we know x is equal to 5 plus y

so now in the other equation i have 3

times 5 plus y plus y

equals 3. now we have an equation with

only one variable this equation

only has the variable y we could solve

this for y

let me distribute the 3 into the

brackets

collect my like terms i've got 15 plus

4y

equals 3. now i want to isolate the y so

i'll move the 15

over 4y equals 3 minus 15 which is

negative 12.

divide both sides by 4 i get y equals

negative 3.

i now need to solve for the other

variable so what we have to do is take

that answer and plug it back into either

of the original equations

it's always easiest if you sub it back

into this rearranged version we have

because

x is isolated so you just take this

answer for y we got

plug it back in for y there and then we

get our answer for x

so i'd have x equals 5 plus y which we

know is negative 3.

so i have x equals five plus negative

three that's five minus three which is

two

so i got x equals two y equals negative

three

same answer again just a different

method example two solve the following

linear system using elimination so this

one's much harder than the first one we

did

for this one we have two equations again

and both equations are already in the

proper format

with x and y on one side of the equation

and the constant on the other side but

we want to eliminate the variables by

adding or subtracting the equations and

to do that

the coefficients have to have the same

absolute value and notice the x's have

different

coefficients and the y's have different

coefficients so what we're going to have

to do is multiply one

or both of the equations by a constant

to make the coefficients have the same

absolute value what sticks out to me is

we can make the x's have a coefficient

of 20

or we can make the y's have a

coefficient of 12 and negative 12. it

doesn't matter let's make the x's both

have a coefficient of 20.

so i would have to multiply equation 1

by 5. so i'm going to do 5 times

equation 1 and that would give me so 5

times 4x is 20x

5 times 3y is 15y and

5 times 13 is 65. make sure you multiply

every term in the equation by the

constant and for the second equation for

the x debit coefficient of 28 i have to

multiply it by

four so four times equation two

and that would give me 20x 4 times

negative 4y is negative 16y

and 4 times negative 7 is negative 28.

now what i'm going to do is i'm going to

figure out should i add or subtract

these to eliminate the variable

the x's have the exact same coefficient

with the exact same sign they're both

positive 20

so subtracting will eliminate the

variable x so let's subtract all the

like terms

20x minus 20x is 0x so that's gone it

eliminates

15y minus negative 16y

is 31y and 65 minus

negative 28 is 93. divide both sides by

31 and we get y

equals 3. now we need to sub y equals 3

into either original equation so i'll

just sub it into equation 1.

let's sub our answer for answer for y

into that equation so

3 times 3 equals 13 and then we just

solve this for x

so 4x plus 9 equals 13.

let's subtract the 9 to the other side i

get 4x equals 4

therefore x must equal 1. so my solution

to this linear system

is x equals one y equals three

if we were to graph both lines they

would intersect at the point one

three example three here's an

application of solving linear systems

a sports shop sells adidas shoes for

eighty-two dollars a pair and

air jensen basketball shoes for 95

dollars a pair one day the shop sells a

combined 75 pairs of shoes

totaling 6 241 in sales how many pairs

of each shoes were sold now for these

types and questions

they're always going to want you to

solve for two variables and in order to

solve for two variables you must be able

to make two equations that involve those

two variables

so i always like to start by figuring

okay what variables am i going to be

trying to solve for

so it wants to know how many pairs of

each shoe are sold so we're trying to

figure okay how many adidas shoes were

sold and how many air jensen shoes were

sold those are my variables

i'll say x equals number of adidas and

y equals number of air jensen in order

to figure out what the value of these

variables are i must be able to make two

equations that involve these two

variables well i know they

sold a combined 75 pairs of shoes so my

first equation

x plus y equals 75. there's one equation

that involves x and y

in the second equation well i know what

the total

dollar amount in sales was and i know

how much each pair costs

adidas shoes cost 82 dollars a pair so

82

times the number of adidas sold plus

air jensen's cost 95 dollars so 95

times the number of air jensen sold

would equal that 6241 dollars when you

have two equations with two variables

you can solve them by

graphing substitution or elimination i

think i'll do elimination for this

let me make the coefficients of the x's

be the same so i'll multiply the first

equation by

82 and that would give me 82x

plus 82y and 82 times 75

is 6150. now let me rewrite equation two

lined up with that right underneath of

it

and since the coefficients of the x have

the exact same sign

i know subtracting will eliminate the

variable 82x minus 82x is 0. 82y minus

95y is negative 13y

and 6150 minus 6241 is negative 91. now

if i divide both sides by negative 13 i

get y equals 7.

so i know that they sold seven pairs of

air jensen shoes

let me just sub that answer back into

either the original equations let me sub

it back into one that's going to be a

lot easier

and then solve for x so i'll sub 7 in

for y

therefore x is 68.

so our final answer for this they sold

68 pairs of adidas and seven pairs of

air jensen

next you need to know how to calculate

the midpoint and distance of a line that

connects two points

so i've got point a and point b it says

calculate the midpoint and the distance

between those points

let's start with midpoint if you notice

the formula it says we need the x and y

coordinates from our first

and our second point now it doesn't

matter which your first point is and

which your second point is but it's

helpful to pick one and label it so i'll

make a my first point which means

5 is my x1 and negative 3 is my y1

it'll make b my second point which means

this is my x2

and this is my y2 right each point has

an x and a y

if i want the midpoint between these two

using the formula

it tells me that what i need to do is

average the x coordinates so i'll add

x1 and x2 together so i'll do 5

plus negative 1 and then i have to

divide it in 2 to get the average of the

two of them

right the middle of the two points is

going to be at the average of the x

coordinates

and the average of the y coordinates so

i now need to average the y coordinates

by adding them and dividing by 2.

so negative 3 plus 5

divided by 2. and the midpoint if i

simplify this

the x coordinate 5 plus negative 1 over

2 that's 4 over 2 which is

2 and the y coordinate negative 3 plus 5

is 2 divided by 2

is 1. let's now find the distance

between point a and b

the distance formula comes from

pythagorean theorem to find the distance

i do the square root of

the difference in the x coordinates so

the formula tells me to do

x 2 minus x 1 and then square it so

negative 1 minus 5

squared plus the difference in the y

coordinates so now i have to do y two

minus y one

so five minus negative three and then

square that difference

and i just have to simplify this so

negative one minus five is negative six

square it and i get thirty six

five minus negative 3 is 8. square it

and i get 64.

so i need the square root of 36 plus 64.

that's the square root of

100 which is 10. so the distance between

the two points is

10 units i don't know what the units are

but that's the distance between the two

points

example 5 says draw the triangle with

these vertices

so let me start by plotting those

vertices so there's my triangle

it says draw the median from c to a b

then find the equation of this median

remember a median of a triangle

is the line segment that joins a vertex

to the midpoint of the opposite side

so the median from c to a b would be the

line that goes from

c to the middle of a b so let me just

estimate where that is right here it

looks like it's going to be about there

that's the middle of a b so the median

would go from c

to that point so what i've drawn in red

is the median let's find the equation of

that line

in the form y equals mx plus b so i'm

going to need to know its slope and y

intercept

let's start by figuring out its slope

and to figure out its slope i would need

to know two points on the line well i

know it's on point c

and it's also at the midpoint of a b

okay so i'm going to need to know the

midpoint of a b

let's find the midpoint so the midpoint

of a b

i'm going to need to average the x

coordinates so negative 1 plus 1

divided by 2 and average the y

coordinates three plus five

over two so the midpoint is going to be

negative one plus one is zero

so zero over two is zero and eight over

two is four

so my midpoint is zero four and i know

point c is on the median as well which

the question tells us is the point three

one

i could use those two points to find the

slope of that median i'll call the slope

m and remember to calculate slope you

just do change in y over change in x y

two minus y one

over x two minus x one so i'll make c my

second point

and the midpoint of a be my first point

so one minus four

over three minus zero and that gives me

negative three

over 3 which is negative 1. so the slope

of my median is negative 1.

if i want the y-intercept of the median

i'm going to have to

into the equation y equals mx plus b i'm

going to have to sub

in the slope i know and a point on the

line doesn't matter which point i can

pick the point zero four

and actually that's already the y

intercept i i know the y intercept is

four but let me just show you

algebraically how we would get it

so we would plug in the point zero four

for x and y the y coordinates four

the x coordinate is zero and we know m

is negative one

and we solve for b four equals

zero plus b so b is four

so now that i have the slope and the y

intercept of the median i could write

the equation

y equals we sub in for m and b so

negative 1x

plus 4 which you would just write as

negative x plus 4.

so that's my final answer that's the

equation of that red line example six

determine the equation for the right

bisector

of the line segment with the endpoints a

and b

let's remember what a right bisector is

the right bisector often called the

perpendicular bisector

is a line that is perpendicular to a

line

and passes through its midpoint so i

know my perpendicular bisector

is going to be perpendicular to the line

connecting a and b so i'm going to need

to know the slope of a

and b which i know i get by doing change

in y over change in x so y

two minus y one six minus negative four

over x two minus x one eight minus

negative two

so that gives me ten over ten

which is one and i know my right

bisector is going to be perpendicular to

this line so it's going to have a slope

perpendicular to one perpendicular

slopes are what are called negative

reciprocals of each other

so i would have to take this one which

you could rewrite as one over one

flip it and change its sign well the

reciprocal of one over one if we flip

that it's still one over one but we do

have to change its sign

so it becomes negative one over one so

negative one

so our slope of our right bisector i'll

call that the perpendicular slope

that equals the negative reciprocal of

one over one which is negative one over

one which is just negative one

if i want the equation of the right

bisector not only do i need it slope but

i need its y

intercept and to find that i'm going to

need to know a point on the right

bisector

well i know the right bisector goes

through the midpoint of a b

so i'm going to need to know the

midpoint of a b so i'll average the x

coordinates

negative two plus eight over two and

average the y coordinates negative four

plus six

over 2. so my midpoint the x coordinate

is 6 over 2 which is 3

the y coordinate would be 2 over 2 which

is 1. so my midpoint

is 3 1 which is an x and a y coordinate

i now take that x and y

and the slope of my right bisector and i

can solve for b the y-intercept

so i'll sub in my point for x and y and

negative 1 for my slope

and if i solve this for b

i get b equals 4. so i can now write my

final equation

by subbing in my m and my b values into

y equals mx plus b

so y equals and make sure you use your

perpendicular slope because we're doing

the right bisector equation so

negative x plus the b value 4.

so there's my final answer example seven

classify the triangle with these

vertices as either scalene isosceles or

equilateral and also state if it has a

right angle

let me start by just drawing it quickly

for you

all right there we go uh it looks like

it's probably scalene

angle d looks like it might be close to

90 degrees

we'll have to do some calculations to

know for sure so if i want to classify

the triangle

and give proof as to what shape it is

i'm going to need to know the length of

each line

which we can use our distance formula

for so let's find the distance of

all three side lengths so let me start

with side d

e now you can do d e or e d doesn't

matter i'll do d

e i'm doing d e i'll make d my first

point

x one y one and e my second point x two

y two

plug that into my distance formula and

get the length of side d e

so i would have the square root of

difference in the x's

squared so negative 2 minus negative 4

squared plus the difference in the y's

squared 6

minus negative 2 squared if i simplify

underneath the square root i'd have

2 squared which is 4 plus 8 squared

which is 64. so 4 plus 64 is 68 so this

would be root

68. we would do the same thing for the

other two sides so first side

ef and df

and then we'll have all three side

lengths for time sake of this video

i'm just going to fast forward through

me solving for these side blanks

okay i have all three side lengths

notice all three sides are different

lengths therefore this is a scalene

triangle

but does it have a right angle well i

know that right angle triangles

the pythagorean theorem is true the sums

of the squares of the shorter two sides

is equal to the square of the longer

side

so let's check is the square root of 68

squared plus the square root of 104

squared is that equal to

the square of the longest side so

root 164 squared if this is true

then there's a right angle if it's not

true then there's no right angle

square rooting and squaring are inverse

operations they cancel each other out so

really what we have here is 68

plus 104 is that equal to

164 and it's not 172

does not equal 164. therefore there's no

right angle therefore

not a right triangle example 8 this is a

really quick one what's the radius of a

circle remember the equation of a circle

is x squared plus y

squared equals r squared in this case r

squared is 36 so the radius

squared is equal to 36 which means the

radius would be equal to the square root

of that

the square root of 36 is 6. we don't

have to consider the negative square

root of it because radius lengths can't

be negative we only look at the positive

so the radius is 6 units

9 for the circle that's centered at the

origin and passes through the point

negative 3

4. so center at the origin passes

through this point negative 3 4. let me

just draw a circle that does that really

quickly for you

that circle uh we want to find the

equation of this circle

and part b asks us to check if this

point 5

2 lies on the circle inside of it or

outside of it and we'll do this

algebraically so if i want the equation

of this circle

well i know one point on the circle this

point right here negative 3

4. that's an x y point the equation of a

circle

is of the form x squared plus y squared

equals

r squared so i need to know what's the

radius squared to figure that out just

plug in your x and y into the equation

so negative 3

squared plus 4 squared

is equal to the radius squared i'd have

9

plus 16 equals r squared

so r squared is 25 which means the

radius is 5 but i don't need the radius

in this question

i just need the radius squared so that i

can write the equation because the

equation is x squared plus y squared

equals r squared

which is 25 so my final equation would

be x squared

plus y squared equals 25.

that's the equation of this circle now

if we want to check if this point lies

inside the circle

outside of it or on the circle we just

plug 52 into the equation of the circle

and see if we get a number that's less

than 25

equal to 25 or greater than 25 that'll

tell us whether it lies inside

outside or on the circle and we can tell

from the graph that the point 5

2 this point right here is clearly

outside of the circle

but if we don't have the graph we want

to prove it algebraically

like i said we take our equation of our

circle and we see what happens when we

plug this value in

so if we plug 5 and 2 in for x and y

well 5 squared plus 2 squared that's

actually going to give me a value that's

bigger than 25 right 25 plus 4 is 29.

29 is bigger than 25. that tells me that

the point must be outside the circle

if we got 25 equals 25 would be on the

circle if we got something less than 25

like if we got 21 less than 25 i mean

inside the circle

example 10 says what's the shortest

distance from point negative 3 5 to this

line y equals a quarter

x plus 10. this question is going to

involve pretty much everything from this

unit

so we've got the line y equals a quarter

x plus 10. i'll just

draw an approximation of that line there

for you there just to visualize what's

happening here

and we have the point negative three

five i don't have a cartesian grid but

let me just put it

right here here's the point negative

three

five i'm looking for the shortest

distance from that point to that line

the shortest distance from the point to

this line isn't going to be something

like that

or like that it's going to be along a

line that's perpendicular to the

original line

so it's going to be along that line like

right there which forms a 90 degree

angle with the original line y equals a

quarter x plus 10 right this black line

is y equals a quarter x

plus 10. and i want to figure out the

shortest distance so basically the

distance from

this point to whatever this point is

here

i don't have that point but i'm

definitely going to need it to find that

point well that's where the blue line

intersects the black line so if i had

the equation of the blue line

i could figure out where they intersect

so to find the equation of the blue line

well i need its slope and y intercept

well i know it's perpendicular to y

equals a quarter x plus 10

so i know its slope is going to be the

negative reciprocal of a quarter

so the perpendicular slope would equal

the negative reciprocal of a quarter

which means flip it and change the sign

so it's negative 4 over 1 which is just

negative 4.

to write the equation of that blue line

i also need the y intercept the b

value to get that into y equals mx plus

b

i have to plug in a point on the line

and the slope well i know a point

negative

3 5 it has an x and a y so sub that in

5 for y i know the slope of the blue

line is negative 4.

i know x is negative 3. if i solve this

equation for b

negative 4 times negative 3 is 12

subtracted over 5 minus 12 is negative

7.

so my b value is negative 7 so the

equation of that blue line

is y equals negative 4 x minus 7. so

this blue line

y equals negative 4x minus 7.

now let's figure out where does that

blue line intersect a black line

i've got two equations i can solve using

substitution or elimination

let me rewrite the two equations

and i think since y is already isolated

in both equations i think substitution

is the easiest method

so really all we do is we take what y is

equal to from one equation

and replace the y on the other equation

with that so replace the y in the second

equation with

a quarter x plus 10.

and now let me solve this equation i

don't really like this

fraction here so what i'm going to do is

i'm going to get rid of the fraction by

multiplying

both sides of the equation by its

denominator so i'll multiply both sides

by 4

and that would give me x plus 40 equals

negative 16x

minus 28. now i'll collect my like terms

onto the same sides of the equation

i've got 17x equals negative 68.

so i've got x equals negative 4.

i now need to figure what y equals so

let me just sub x equals negative 4 into

either original i'll just plug it into

the first one because i have room

underneath it there

so i'll plug negative 4 in for x

a quarter times negative four is

negative one so i've got negative one

plus ten which is nine

so my point of intersection is the point

negative four nine so i know this point

right here is the point

negative four nine the shortest

distance is going to be the distance

between negative 3 5

and negative 4 9.

so if i use the distance formula i'll

have my final answer

so i'll do the difference in the x

coordinates squared plus the difference

in the y coordinates

squared so this gives me

1 plus 16 so i have

root 17

units so that's the shortest distance

from the point to the line

[Music]

in this unit you look at the three

versions of equations a quadratic

function can be given to you in

standard vertex and factored form you

learn the usefulnesses of all three

versions

and how to transition between the three

versions now quadratics

are equations where the the highest

degree exponent on x is a two

and the shape of those equations is

always this u-type shape which we call a

parabola

now a standard form equation is most

useful for determining the y-intercept

so the y-intercept we can tell from the

equation because it's the c value

so this point right here would be the

point 0 and then whatever the c

value is in the standard form equation

another useful thing about standard form

is that the a value tells you the

direction of opening

if a is positive the parabola opens up

like the one we have here

if a is negative it opens down

now a quadratic function could also be

written in vertex form or factored form

and those have different uses now the a

value in all three versions of the

equation all tell you the same thing

they all tell you direction of opening

so let's just talk about what the other

variables tell you so in vertex form

well it's obviously most useful for

telling you what the coordinates of the

vertex are

so if it's in vertex form y equals a

times x minus h squared plus k

the vertex is at h k

and notice since the general format is x

minus h this h value is always going to

be the opposite sign of what you see in

the equation

so our vertex is at h k

that's what vertex form is useful for

and factored form is useful for the

x-intercepts

so these points right here you could

figure those out by if we have it in

factored form the form y equals a times

x minus or times

x minus s we can find the x intercepts

by looking at the r

and s values this x intercept i could

say is at

r0 and this one is at s0

so the x-intercepts all right

r and

s let's look at how we can work with the

three versions of a quadratic and how to

transition between the three

question 11. this quadratic is in vertex

form

complete the table of information so the

vertex is going to be at negative 4

3 right because the general format is x

minus h so it must be x minus negative 4

in order for it to look like x plus 4

which means h is actually negative 4.

so my vertex is at negative 4 3.

x coordinates always opposite sign of

what you see there

the axis of symmetry is the vertical

line that would divide the parabola in

half

and that vertical line notice it passes

right through the vertex

so the equation of the axis of symmetry

is x equals whatever the x coordinate of

the vertex is so x

equals h so in this case x equals

negative 4. stretch your compression we

get that by looking at the absolute

value of a

so the absolute value of negative 2 is

2. so if the absolute value of a

is bigger than 1 it causes a vertical

stretch so vertical

stretch we say by a factor of the

absolute value of a so the absolute

value of negative two

is two always use the absolute value

when describing the stretch factor

the direction of opening is affected by

the sign since the a value is actually

negative that tells us the quadratic is

going to open

down values the x may take

also called the domain of the function

the answer for quadratics is always the

same there's no restrictions on the

domain we just say

x can be any real number

values that y may take well since the

parabola opens down

we know that the vertex is going to be

at the very top of the parabola right

the vert the parabola is going to look

like this the vertex is at the top

so we know the y values are always going

to be at or below the y coordinate of

the vertex

so the y values are always going to be

less than or equal to the y coordinate

the vertex which is 3.

if i'm going to graph this function i'm

going to want to make a table of values

and in the table of values you always

want to put the vertex in the center so

i'm going to put negative 4

3 in the center and now i want to pick

points on either side of the vertex so

to the left of negative 4

i'm going to choose x values negative 5

and negative 6

and then to the right of negative 4 i'll

choose x values negative 3 and negative

2.

and now i just have to plug those in to

my equation

for x and calculate the y's so if i plug

negative 6

in for x i would have y equals negative

2

times negative 6 plus four squared

plus three if i evaluate this negative

two squared is four

times negative two is negative eight

plus three is negative five

so this would be negative five and since

i know parabolas are symmetrical if two

to the left of the vertex the y

coordinates negative 5 2 to the right

it's going to be at the same spot

let me plug negative 5 in for x and see

what i would get negative 5

plus 4 is negative 1 squared is 1 times

negative 2 is negative 2 plus 3

is 1. so i know those are both 1. so

using those points

i could graph the quadratic

and there's a rough sketch of the

parabola let's go the other way here's

the graph let's find the vertex form

equation so remember vertex form looks

like this

a times x minus h squared don't forget

the squared

plus k in order to write the equation

well i just need to know the vertex

plug that in for h and k but i also need

to know a the stretcher compression

factor

and we'll solve for that by plugging in

the vertex and a point x y

so i've got the vertex here it's at 2

negative 6

and it gives me another point here it

labels the y intercept that's at 0

negative 4.

i've got my x y my

h k sub all that in and solve for

a i've got y value of negative four i

don't know a that's what we're solving

for

x is zero h is two squared

plus my k value of negative six so plus

negative six i'll just write minus six

let's solve this for a so in the

brackets i'll do first zero minus

two is negative two square that i get

four

now it often helps instead of writing a

times four usually people write that as

4a if we write it like that you'll avoid

mistakes

let me know add the 6 over negative 4

plus 6 is 2. so i have 2 equals 4a

divide both sides by 4 i get a half

equals my a value so my final answer

would have a

h and k plugged into the vertex form so

a half

x minus 2 squared minus 6.

there's the vertex form equation 13 just

asked me to describe

transformations in words so we can

describe the left and right shift based

on where the vertex is so my vertex

is at negative 5 negative 4.

i get that from the h and k remember h

is always the opposite sign of what we

see there

the vertex of x squared is at zero zero

but now the vertex of

this function is at negative five

negative four so for that to happen it

must have shifted left

five units and shifted down four units

now if i look at the absolute value of a

that will tell me the

vertical stretch or compression factor

the absolute value of a is a third

if the absolute value is between 0 and 1

that's going to vertically compress

it so i would say vertical compression

by a factor of the absolute value of a

so

by a factor of a third but the fact that

it's negative means it's also going to

be opening down which means it's been

vertically reflected so we would say

vertical reflection example 14.

for this quadratic that's in factored

form we want to state the x-intercepts

the vertex and the axis of symmetry

and use the information to graph it well

to find the x-intercepts if it's in

factored form that's easy we're just

looking to see

what value of x would make any of the

factors be 0.

what would make x minus 2 be 0 would be

if x was 2

so 2 is an x intercept so we have an x

intercept

at 2 0 and what value of x would make

this factor be 0 well 4 would make it be

0.

so there's another x-intercept at 4 0.

right if we can make one of the factors

be 0 we make the whole product be 0

which makes the y-coordinate become 0.

so 2

and 4 are both x-values that have a

y-value of 0

meaning their x-intercepts and that's

easy to pick out from factored form

right it's just these numbers here

remember opposite sign of what you see

there now where's the vertex well i know

parabolas are symmetrical so the vertex

is going to be halfway between those

so a shortcut to find the x-coordinate

of the vertex

would be to just find the average of the

x-intercepts so the x-coordinate of the

vertex would be equal to the average of

two and four

and you average numbers by adding them

dividing by two so the x-coordinate of

the vertex is three

but where's the y coordinate to find the

y coordinate of the vertex

we have to take that x coordinate and

plug it into

the original equation so i would have

negative 4

times 3 minus 2 times 3 minus 4

which is negative 4 times 1 times

negative 1

which is 4. so my vertex is at the point

3 4. and the axis of symmetry

is the vertical line that goes through

the middle of the parabola so the

parabola looks let's do a rough sketch

now the parabola looks something like

this

the axis of symmetry would be the

vertical line and it's going to go right

through the vertex

the equation of that line would be x

equals

the x-coordinate of the vertex so x

equals 3 is the equation of the axis of

symmetry

15 a parabola has x-intercepts of

negative 8 and 2 and passes through the

point 0 negative 8 determine the

equation of this parabola in the form

here this factored form okay well the

intercepts

are our r and our s and we know a point

x y to write the equation in factored

form we just have to solve for a

so what's sub in what we know y is

negative eight

i don't know a but my x value

is 0 so i do 0 minus r so minus negative

8

times 0 minus s and s is 2.

now let's solve this for a negative 8

equals a 0

minus negative 8 is eight and zero minus

two is negative two

so negative eight equals a times eight

times negative two

i'll write that as negative 16 a divide

both sides by negative 16 to isolate a

and i get a equals a half

so i could write my final equation y

equals a half

and then sub in for r and s but leave x

and y so

x minus negative eight which we would

write as plus eight

and x minus 2.

there we have it question 16 factor each

of the following

part a i have a three term quadratic

the coefficient of that x squared term

is one which means i can do product and

sum factoring the short way so i just

have to find numbers who

have a product of the c value

24 and a sum of the b

value 10. the numbers that multiply to

24 and add to 10 are 6

and 4. since the coefficient of the x

square is 1 i can go right to my

factored form equation

by just

adding 6 and 4 to x in both of those

factors

so x plus 6 and x plus 4

are my factors part b is very similar 3

term quadratic coefficient of x squared

is 1.

so i just need to find numbers that have

a product of the c value negative 12

and a sum of the b value which is one

and the numbers that satisfy that

product and sum are four

and negative three so once again what i

do

is i can go right to my factored form

equation by adding those two numbers to

the x's in each of these factors

so x plus 4 and x plus negative 3 which

we would just write as x minus 3.

part c 3 term quadratic but the leading

coefficient the coefficient of the x

squared is not 1. so what we should do

is we should check if it can common

factor out

and it can in this case two divides

evenly into

22 and 48 so let's common factor of the

two by dividing all three terms by 2 and

putting it out front so x squared

plus 11x plus 24 is my second factor

and now we're just going to look inside

the brackets here at this quadratic and

we're going to try and factor that

the coefficient of the x squared is now

one so we're just looking for numbers

that have a product of the c value 24

and a sum of the b value 11. the numbers

that work

are eight and three

so we can factor what's in the brackets

to x plus eight times x plus three

just don't lose that two that was out

front at the beginning

part d a three term quadratic that we

want to factor

the coefficient of the x squared is not

one it's two but this time we can't

common factor out of 2 2 doesn't divide

evenly into 7

and negative 15. so we can't common

factor it out so we have to factor this

the long way

some teachers may call it by grouping

some by decomposition or just the long

factoring method

but whatever you call it we're looking

for numbers that don't just have a

product of the c

value now a product of a times c so

product of 2 times negative 15 is

negative 30

and a sum of the b value 7. so the

numbers that work in this case

are 10 and negative 3. and since the

leading coefficient is not

1 we can't go right to our factors what

we need to do is

split the middle term into 10x minus 3x

2x squared plus 10x minus 3x

minus 15. so i rewrote the equation but

i rewrote 7x as 10x minus 3x

and specifically those two numbers

because those are the numbers that

satisfy the product and sum

and then we factor it by grouping we

look at the first two terms and take out

a common factor from the first two so i

could take out a 2x

and that would leave me with x plus five

and then i look at the last two terms

negative 3x minus 15 and i take a common

factor from those last two terms

i could take a negative 3 from both

terms and that would leave me with once

again

x plus 5. and now that we have a common

binomial they both have that x

plus 5 we can factor out that x plus 5

and be left with 2x minus 3.

e once again a quadratic leading

coefficient is not 1

and you can't common factor it out 3

doesn't divide evenly into 23 and

negative 8.

so we're going to have to factor it the

long way by decomposition and grouping

so i want a product of a times c

3 times negative 8 is negative 24 and a

sum of the b

value which is 23 the numbers that

satisfy that product and sum are 24

and negative 1. so i need to split the

middle term

into 24x minus 1x

and now i factor by grouping if i look

at the first two terms and take out a

common factor i can take out a 3x and be

left with x

plus 8. the last two terms i take out a

negative one

and i'm left with x plus eight i see

that common binomial that's how i know

i'm doing it right

take out that common binomial and you're

left with three x

minus one as your second factor f is a

special product it's a difference of

squares

i have an x squared minus

a 16 and 16 is a perfect square number

it's a four squared

so if you have a difference of two

perfect square numbers

it factors to x minus four

times x plus 4.

g is another difference of squares i

have 4

x squared minus 25 well this one's a

little trickier because in order to

think of the first term as a perfect

square number

we'll think of it as a 2x that's being

squared right if we square the 2 and the

x we get

4x squared minus 25

is 5 squared and now that i see it's a

difference of squares

i know it factors to 2x minus 5 times 2x

plus 5.

part h is actually what we call a

perfect square trinomial

and that's because it's the same number

twice that satisfies the product and sum

the coefficient of this is one so i just

need to find numbers with product of c

which is nine and a sum of b which is

six and the numbers that work are

three and three so this would factor

to x plus three times another x

plus three which you would never write

it twice

you would just write it as x plus 3

squared

let's go backwards now instead of

factoring let's expand example 17

says expand each of the following into

standard form so i'm going to have to do

my double distributive property

sometimes called foiling

multiply the first terms the outside

terms

the inside terms and the last terms

that's why it's called foiling

and if i write all four of those

products so 4x times

x is 4x squared and then 4x

times 7 is 28x negative 1 times x

is negative x and negative 1 times 7 is

negative 7.

collect my like terms my only like terms

are 28x minus 1x which is positive 27x

expanding part b is probably the most

common mistake people make in the grade

10 math course

x minus 5 all squared is not just x

squared minus 5 squared

you can't put the squared on both terms

you actually have to foil it out you

have to do x minus 5 times another x

minus 5.

rewrite it like that and then do your

double distributive property

and when you do all four products you'll

see what you get

x times x is x squared x times negative

5 is negative 5x

then negative 5 times x is another

negative 5x and then negative 5 times

negative 5 is positive 25.

if i collect my like terms the middle

two terms negative 5x minus 5x is

negative 10x

and there it is expanded okay question

18 i want to get these standard form

quadratic equations

into vertex form quadratic equations the

process for doing that

is you start by putting the first two

terms in brackets

and then you common factor out the

coefficient of the x squared from the

first two terms but it's just one in

this first example so i don't have to do

anything

after that what we do is we add and

subtract

a special number inside the brackets so

i need to add

and subtract a certain number inside the

brackets here

now what number goes here and here

you always do half of this number and

then square it

and then put that in these two spots so

half of 6 is 3.

square it and we get 9. so we add 9

and subtract 9. really we added 0 right

9 minus 9 is 0. but that's the step

that creates a perfect square trinomial

which is going to help it factor very

nicely into vertex form

before i can factor it though i need to

get this negative nine out of the

brackets i don't really want it in there

so you have to get it out by multiplying

by whatever's in front of the brackets

which in this case is just one so really

we're just moving the negative nine out

so i have x

squared plus six x plus 9

negative 9 moves out now i just have two

things left to do

factor this quadratic so what numbers

multiply to 9

and add to 6 are 3 and 3 so it goes to x

plus 3 times another x plus 3 which is x

plus 3

squared we very intentionally created it

so that it would do that

and outside the brackets negative 9 plus

11 is positive

2. so my vertex is negative 3

2 and since the a value is positive 1 i

know the parabola opens

up which means the vertex is at the

bottom so the vertex is emit

so the vertex is a min point let's do

the question again but

part b a little more difficult this time

same steps put the first two terms in

brackets

step two whatever the coefficient of the

x squared is common factor it out from

the first two terms that are in brackets

so i'll put the three out front

divide both of the terms and brackets by

three i get x squared plus eight x

and now what we do is inside the

brackets same thing we want to

create a perfect square trinomial by

adding and subtracting

half of this eight squared right you

always look at this number take half of

it and then square it

so half of 8 is 4 4 squared is 16. so

i'm going to add and subtract 16

but we don't want that negative 16 in

there so i take it out of the brackets

by multiplying it by

whatever's in front so i have y equals

three

x squared plus eight x plus sixteen

outside of the brackets this this

negative sixteen is being multiplied by

three so it's negative forty eight

and then i just have to factor the

quadratic the numbers that multiply to

16 and add to 8 are

4 and 4 so it factors to x plus 4

times another x plus 4 which we write as

x plus 4 squared

and then if i do negative 48 minus 17 i

get negative 65.

so my vertex is negative 4

negative 65 and because the a value is

positive i know the parabola opens up

which means the vertex is at the bottom

so once again it's a min point

almost done with quadratics let's just

solve some equations

so when solving quadratic equations you

always want to get it into factored form

set each factor to zero and solve each

of those equations you make

so the first one's a difference of

squares it's an x

squared minus a six squared so i know

that factors to

x minus six times x plus six

and how can this equation be true is if

either of the factors were 0 it would

make the whole product be 0 making the

whole equation true

so set each factor to 0 so x minus 6

equal to 0

set x plus 6 equal to 0 and solve both

of those equations you made

so i get a first answer x is 6

and my second answer is x equals

negative 6.

these are both answers that satisfy the

original equation this one

this one's not set to 0 to start with so

we should always start by doing that if

you have a quadratic start by setting it

to zero so x squared plus four x

minus 21 equals zero now we want to

factor it so i'd be looking for numbers

that have a product of negative 21

and a sum of four and since the

coefficient of the x squared is 1

i'm going to be able to go right to my

factors once i find the numbers that

satisfy that product and sum so the

numbers that satisfy the product and sum

are 7 and negative 3. so they go in

those two spots

and now to find how the product of these

two things could be zero is if either of

those two things were equal to zero

so set both the factors equal to zero

and solve both of those equations my

first answer is negative seven

my second answer three see

it's set to zero the coefficient of the

x squared isn't one it's negative one

which i don't really like so i'm going

to common factor that out so divide

all three terms by negative one it just

changes the sign of all of them

and now in the brackets is a quadratic

with a leading coefficient of one so i

would just need to find the numbers that

have a product of negative six and a sum

of negative five and put those numbers

added to x

so those numbers would be negative six

and one

and now this product is zero if any of

the factors are zero

so set the factors that involve an x

equal to zero

and solve each of those equations

and solve for those two cases so i get

six

and negative one for my two answers this

one

start by setting it equal to zero i

can't common factor with that five so

i'm going to have to factor this the

long way

the numbers that multiply to five times

negative 4 so i have a product

of negative 20 and a sum of the b value

negative 19 well that'd be negative 20

and 1.

so i'll split the middle term into

negative 20x

plus 1x and now i'll take a common

factor from the first two terms i'd take

out a 5x and i'd have x minus 4. from

the last two terms i can just take out a

1

which leaves me with x minus 4. i have

that common binomial of x minus 4 when i

take that out i'm left with 5x

plus 1. and now i set each factor to 0

and solve each of those equations i get

4

and this one's a 2-step one to solve

subtract the 1 over and then divide the

five

my second answer is negative one over

five

two more to solve notice the first four

i solved by factoring

these ones actually aren't factorable

there's nothing that multiplies to five

and adds to seven so we can't factor it

but that doesn't mean there's no

solutions it just means there's no

rational solutions there may be some

irrational ones so we need quadratic

formula to check

so if i use quadratic formula x equals

negative b so negative 7 plus or minus

the square root

of b squared so 7 squared minus 4

times a times c and this is

all over 2 times a

now i recommend to simplify underneath

the square root first

because we call that the discriminant

and that's going to reveal to us how

many answers we're going to get

so 7 squared is 49 minus 20

that's 29. so i have root 29 okay since

i got a positive

discriminant that means i'm going to get

two answers here

so i'll split them into my two answers

i've got negative 7

plus root 29 over 2

and i also have x equals negative seven

minus root 29

over two okay i've got my two exact

answers let's get to approximate answers

by evaluating on our calculator and

rounding to two decimal places

the first one would give me negative

zero point eight one

and the second one would give me

negative 6.19

if we look at f i mean sometimes we

can't get any answers you'll see what i

mean with this question

this one not factorable there's no

numbers that have a product of

14 and a sum of four so we can't factor

it so we try quadratic formula

x equals negative b so negative 4 plus

or minus

the square root of b squared

minus 4 times a times c

all over 2 times a the discriminant

if we simplify that 16 minus 56

we actually get a negative answer we get

negative 40.

so we've got the square root of negative

40. and you can't square root a negative

so this means we're not going to get any

real solutions for this so we'd say no

real solutions

we're almost through the whole course 20

sketch a graph and label all key

properties of x squared plus 8x plus 12.

so key properties would involve

x-intercepts vertex

y-intercept well from standard form i

know the y-intercept

it's the constant value 12. so i'll plot

that right now i know my y-intercept

i'm also going to want the x-intercepts

so i'm going to get this into factored

form

the x-intercepts have a y-coordinate of

zero so i'm going to set y to 0

and now i'm going to solve this and then

that will give me the x intercepts

so to solve it i'll have to factor it

what numbers have a product of 12 and a

sum of 8

those numbers are 6 and

2. so there's factored form the product

would be zero if either of the factors

were zero

so set each factor to zero and solve and

i get

negative six and negative two so my

x-intercepts are

at negative six and negative 2.

let me find the vertex now well i don't

have to worry about putting this

quadratic into vertex form

because i know the vertex is going to be

halfway between the x-intercepts because

parabolas are symmetrical so i can find

the x-coordinate of the vertex

just by finding the average of the

x-intercepts so add them

and divide by two negative eight over

two is negative four the vertex is going

to have an x-coordinate of negative four

what's the y-coordinate of the vertex

just plug negative 4 into the original

equation

so be negative 4 squared plus 8 times

negative 4

plus 12. 16

minus 32 plus 12. hey that's negative 4

as well so my vertex is at negative 4

negative 4.

so that's this point right here i might

as well plot one more point just

because i know parabolas are symmetrical

if four units to the right of the vertex

write a y coordinate of 12 4 units to

the left 1 2

3 4 we would also be at a y coordinate

of 12.

so now i should be able to draw a fairly

accurate sketch of this parabola

making sure i go through all of these

key points here

and have that good u shape and there we

go last quadratics question is an

application question

an object is launched upward at 64 feet

per second from a platform 80 feet high

the equation for the object's height and

feet is based on time in seconds is

given by this equation

when does the object land on the ground

anything that time you see anything

about the ground that's the x-intercepts

that's when the height is zero so set

the height to zero

and solve and to solve a quadratic

that's set to zero we factor it

so i'm actually going to common factor

out this negative 16 first because i

think it divides evenly into 64 and 80.

so let's divide all three terms by

negative 16

and i would get x squared minus 4x

minus 5. and now this quadratic has a

leading coefficient of 1 so i just have

to find numbers that multiply to

negative 5

and add to negative 4 and those numbers

are negative 5

and positive 1. so there's my factored

form version of it

and now this whole product would be 0 if

any of the factors containing an x

was 0. so set each factor with an x to 0

and solve i would get 5

and negative 1. now keep in mind x

in this equation actually stands for

time in seconds we can't have negative

time so we can reject that answer

and our only answer here is 5 seconds so

it'll hit the ground after 5 seconds

part b says what is the max height of

the object

well the max anytime you see anything

about max or min it's talking about the

vertex

and what is the max height it wants to

know the y-coordinate of the vertex

that's we're looking for here

it's going to be easier for us to find

the x-coordinate first and then use that

to find the max height

so i can find the x-coordinate of the

vertex by finding the average of the

x-intercepts

so the x-coordinate of the vertex i know

it's going to be halfway between

the x-intercepts so i just have to add

them negative 1 and 5

and divide by 2. that gives me 4 over 2

which is

2. so the x coordinate of the vertex is

2 seconds so it's going to be at a max

at 2 seconds but what is the max height

well let's find the y coordinate of the

vertex

by plugging 2 into the original equation

if i plug it into the original negative

16 times 2 squared

plus 64 times 2 plus 80.

and if i evaluate this i would get 144

feet so my vertex is at the point 2 144

the x coordinate is the time

the y coordinate is the height so the

max height is 144 feet

part c says when is the object 100 feet

off the ground

so this one we set the original equation

we set the height to 100

and then we need to solve this equation

so in order to solve a quadratic we have

to set it to zero so let's move that 100

to the other side

80 minus 100 is negative 20.

and now to solve this i can't common

factor out that entire negative 16

because that doesn't divide evenly into

20 but i could common factor out

part of it so 0 equals i could take out

let's say a negative 4 from all three

terms so i'll take our negative 4

and that would give me 4x squared minus

16x

plus 5. and now

are there any numbers that have a

product of 4 times 5 so a product of 20

and a sum of negative 16 i don't think

so so we'd have to use quadratic formula

to see how this factor could be zero

so quadratic formula with that factor i

would do

x equals negative b so 16

plus or minus the square root of

negative 16

squared minus 4 times a

times c all over

2 times a notice this negative 4 is not

part of the quadratic equation at all

all we're doing is trying to figure out

how can this be 0 because that will make

the whole product be 0 which is what we

want

and then if we simplify the discriminant

underneath the square root

i would get 176

and if i split this into my two answers

if i do 16 plus root 176

over 8 i get 3.66

and if i do 16 minus root 176 over 8 i

get

0.33 i get two positive answers

so i get two times when it's at 100 feet

and if we think about if we think about

what it looks like we know it has a

y-intercept of 80 that's how high the

platform is

and then of course it's going to go up

and come back down the highest it gets

we figured out was 144 this was at 2 144

and this is at 0 80.

so it must be at a height of 100 at two

other times

at 0.34 and at 3.66 and that's what we

found

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in this unit you start off by looking at

similar triangles which

are triangles that have the same shape

but are different sizes

we know similar triangles have

equivalent ratios of sides

that naturally extends to right angle

trigonometry

and the acronym sohcahtoa which tells us

about ratios of sides in

similar right angle triangles we look at

this right angle triangle here

we know that if a right angle triangle

has a reference angle of theta

all other right angle triangles with

that same angle theta

are similar they're the same shape but

they may be different sizes

but because they're similar we know they

have equivalent ratios of corresponding

sides

since a triangle has three sides there's

three different pairs we could make from

those three sides we could pair up

opposite and hypotenuse to get a ratio

and we call

that ratio sine that's where so comes

from sine of an angle equals opposite

over hypotenuse

if we pair up the adjacent and

hypotenuse we call that ratio

cosine cosine of angle is adjacent over

hypotenuse

and tan of an angle is opposite over

adjacent and

the sides are being labeled from this

reference angle

theta so of course opposite from that

would be that side adjacent

means right beside it which is right

there the hypotenuse is always across

from the right

angle so here's a diagram just

illustrating that notice

this right angle triangle i have the

sine cosine and tan ratios being

calculated

for this 40 degree angle notice if i

keep the shape of this the same so

keep that 90 degree angle in that 40

degree reference angle but change the

size of this triangle

notice that these three ratios are all

going to stay exactly the same so even

though the size of the triangle changes

because all these triangles are similar

their ratios stay the same

and your calculator has the ratios

programmed in so sohcahtoa is used for

right angle triangles so let's do a

couple questions where we're going to

have to solve for angles or sides in

right angle triangles

so i've got this right angle triangle

here i see a reference angle 41 degrees

and it's asking me to solve for the

indicated side indicated by x so i need

the length of side x

and i know this side here so from the 41

degree angle

i want the side that is opposite to it

and i know the hypotenuse so what ratio

has opposite and hypotenuse that's sine

so i know

sine of the 41 degree angle

would equal the opposite side which we

have called x over the hypotenuse which

is 30.

and then we'll just isolate x to solve

for it by multiplying the 30 over

so 30 multiplied by sine of 41 degrees

equals x and if we evaluate that 30

times sine 41 making sure your

calculator is in degree mode

we'll get an approximate answer for the

length of side x and it's about 19.68

and our units for this are meters

part b once again we have a right angle

triangle where we're looking for a side

length so we can use sohcahtoa

here's our reference angle from that

reference angle we know the opposite

side and we are looking for the adjacent

side

the ratio that has opposite and adjacent

is tan so i know

tan from the reference angle

would equal opposite which is 8 over

adjacent which is x

now when the variable's in the

denominator of the ratio when you do the

algebra to isolate it really what

happens is the x and the tan 27 just

switch spots

so i get x equals 8 divided by

tan of 27 degrees and i can get my

approximate value for that by typing it

on my calculator

and i get 15.7 meters

this one we're looking for an angle when

we have a right triangle and we know two

sides and we want an

angle we actually end up using inverse

trig ratios

let me show you so from the angle we

want we know the opposite side

and we know the adjacent side so the

ratio that has

opposite and adjacent is tan so i know

tan of that angle

would equal opposite 11 over adjacent

seven to get the angle i actually have

to do inverse tan

so theta is equal to inverse tan

which we denote as tan with that

negative one looking exponent thing but

it's not actually an exponent just means

inverse tan

of the ratio 11 over seven

and so if we evaluate this inverse tan

of 11 over seven

when you use inverse tan the calculator

knows your input is the ratio and it's

going to output the angle

we do that we'll get an angle of about

57.53 degrees if we round it to two

decimal places

let's try another one like that once

again a right angle triangle

we know two sides we're looking for an

angle so from the angle we want

we know the opposite side and the

hypotenuse

the ratio that has opposite and

hypotenuse is sine so i know

sine from that angle would equal 51

over 54. and when we want the angle and

we know the ratio

we do inverse sine of the ratio

and the calculator will output to us the

angle that has that ratio

and it'll be about 70.81 degrees

let's move on to non-right angle

trigonometry

so after you learned about sohcahtoa you

would have then learned about

two laws that work for calculating sides

and angles of non-right angle triangles

called oblique triangles and you would

have learned about sine law and cosine

law

now sine law and cosine law also would

have worked for these first four but

sohcahtoa is so much quicker

that if you have a right triangle you do

sohcahtoa if you have a non-right

triangle

sohcahtoa does not work you have to use

sine law and cosine law

so depending on what you're given and

what you're looking for you decide what

to use so for this question

we know two angles and one side of the

triangle

which means we're going to use sign law

to find the other side we're looking for

side a which is of course across from

angle a

sine law tells us that the ratio of a

side to the sine of its opposite angle

is equal for each of the sides and its

corresponding opposite angle

so i know that a over sine

of its opposite angle 54 would be equal

to

13 divided by sine of its opposite angle

67

so based on this i can now i now have an

equation with one variable i can just

multiply this sine 54 to the other side

and i then have

a isolated so 13 times sine 54

over sine 67 if i evaluate this

i'll get an approximate length for side

a which is 11.43 centimeters

part f if we're looking for an angle and

we know all three sides

we can actually use cosine law for this

so if i want cosine

of angle p it's equal to we do the

square of the side

opposite from the angle we want so 6

squared

minus the squares of the other two sides

so minus seven squared minus five

squared

the order of the seven and the five

doesn't matter but the six definitely

has to be first the first one is the one

across from the angle we want

divided by negative two times the second

two sides the seven and the 5.

okay so we have the ratio maybe we want

to simplify this ratio

so 6 squared minus 7 squared minus 5

squared is negative 38

and then negative 2 times 7 times 5 is

negative 70.

so i have the ratio and i want the angle

i know to get the angle if i want angle

p

i can do inverse cosine of the ratio

and i'll just cancel out those two

negatives and just write 38 over 70.

so angle p if i do inverse cos of that

ratio i get 57.12

degrees and the last scenario you'll

have is if you know

two sides and the angle contained by

those two sides

which means between the two sides you

can find the side opposite from the

contained angle

by doing cosine law a rearranged version

of cosine law compared to this one we

did to find the angle

but it's the same equation so if i want

this side

s equation based on cosine law is s

squared

equals the sum of the squares of the

other two sides

7 squared plus 10 squared minus 2

times those two sides we know 7 and 10

times cosine of the angle contained by

those two sides

54. so let me simplify the right side of

this equation

49 plus 100 is 149 minus

2 times 7 times 10 is 140

cos 54. now let me get an approximate

value for the right side

i get 66.71

zero zero six four six

eight that's what s squared is equal to

i kept all those decimal places because

we couldn't we shouldn't round till the

end

s would be approximately equal to the

square root of this

and if i square with that rounded to two

decimal places it's about 8.17

kilometers all right so that's it for

trigonometry if you have a right angle

triangle use sohcahtoa if you have an

oblique triangle which means a non-right

angle triangle you need to use sine law

and cosine law

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