Question

http://prntscr.com/63nfos

Answer 1

@johnweldon1993

Answer 2

SOOOO what type of triangle is this?

Answer 3

A triangle ;)

Answer 4

Just kidding, 90 degrees? o-o

Answer 5

Lol SOOO correct ;P

And yes, it has a right angle, so this is a right triangle.

That is the clue that we CAN use cos to solve for that angle

Answer 6

Oh yay, I was right? :P Awesome.
And okay, what would I do next?

Answer 7

So, what does cos mean?

What 2 sides of a right triangle are we interested in when we use the cos function?

Answer 8

Well.. I have some guesses, can you see if I'm right?

Answer 9

Absolutely :)

Answer 10

\[PR=\sqrt{12^2 +20^2}\]

Answer 11

How's that?

Answer 12

While that IS a true statement...that would be how we find the length of that hypotenuse...but gives us nothing in terms of the angle

Answer 13

oh yeah, that's right.

Answer 14

So, in terms of the angle...do you know anything about the trig function 'cosine' ?

Answer 15

yeah, I'm a bit familiar with that..

Answer 16

Have you seen

SOH CAH TOA before?

Answer 17

It looks familiar >_< what do they stand for?

Answer 18

It is a clever way to remember the trig functions and how to use them

S "sin"
O "opposite"
H "hypotenuse"

C "cosine"
A "adjacent"
H "hypotenuse"

T "tangent"
O "opposite"
A "adjacent"

So, Think of it as

Some Old Horse, Caught Another Horse, Tripping On Acid

or some other fancy clever anagram lol

But really just remember

\[\large sin(\theta) = \frac{opposite}{hypotenuse}\]
\[\large cos(\theta) = \frac{adjacent}{hypotenuse}\]
\[\large tan(\theta) = \frac{opposite}{adjacent}\]

Answer 19

very helpful, thank you :)

Answer 20

so we could you CAH? right?

Answer 21

Indeed!

So that says that

"cosine" is adjacent over the hypotenuse

So

\[\large cos(\theta) = \frac{?}{?}\]

Answer 22

\[\frac{ RQ }{ PR} ?\]

Answer 23

Indeed

Now, just replace them with the values :D

*You already found the hypotenuse above so this is perfect*

Answer 24

okay i got 0.514

Answer 25

Awesome, thanks :)

Answer 26

Of course :)