Question

Find the measure of angle Q in triangle below

Answer 1

there

Answer 2

now, that you have the law of cosines formula, can you factor it in a way that you'd ISOLATE the cos() part?

Answer 3

So you just plug in 48 and 36 for the lengths. But I don't know the angle....

Answer 4

I dont know any of the angles... do I use the sides to tell me?

Answer 5

well, lemme retype the formula a bit

Answer 6

$$
q^2=\color{blue}{48}^2+\color{red}{36}^2-2(\color{blue}{24}\times\color{red}{36})cos(Q)\\
q=\sqrt{\color{blue}{48}^2+\color{red}{36}^2-2(\color{blue}{48}\times\color{red}{36})cos(Q)}
$$

Answer 7

so, can you factor it in a way that you'd ISOLATE the cos() part?

Answer 8

...the sides?

Answer 9

well, factor it so it ends up looking like

cos(Q)= 48+....x.... and so on

Answer 10

the side FACING OFF angle Q is 60, so the formula, adding that would be
$$
\color{green}{60}^2=\color{blue}{48}^2+\color{red}{36}^2-2(\color{blue}{24}\times\color{red}{36})cos(Q)\\
\color{green}{60}=\sqrt{\color{blue}{48}^2+\color{red}{36}^2-2(\color{blue}{48}\times\color{red}{36})cos(Q)}
$$

Answer 11

once you factor it out to isolate the cos() part, you'd get your angle

Answer 12

OH

Answer 13

12?

Answer 14

it'd be like
cos(Q)= something plus something square and such
\(cos^{-1}[cos(Q)]= cos^{-1}[something plus something square and such]
Q = THAT value, make sure your calculator is in Degrees mode :)

Answer 15

12? lemme check, based on that picture, it looks bigger than 12 :)

Answer 16

12 degrees that is

Answer 17

Yay

Answer 18

ahem is not 12

Answer 19

Oh. okay lemme recheck

Answer 20

if you look at the picture, is much bigger than 12 degrees

Answer 21

Oh yeah, it is

Answer 22

lemme check

Answer 23

It looks like 100+

Answer 24

degrees

Answer 25

ahemm, is not 12

Answer 26

Yeah, I know it s not

Answer 27

oh, ok, lemme factor the formula

Answer 28

THen where am I messing up?Cause i plugged it in

Answer 29

$$
\color{green}{60}^2=\color{blue}{48}^2+\color{red}{36}^2-2(\color{blue}{24}\times\color{red}{36})cos(Q)\\
\color{green}{60}^2-(\color{blue}{48}^2+\color{red}{36}^2)=-2(\color{blue}{24}\times\color{red}{36})cos(Q)\\
----------------------------\\
\cfrac{\color{green}{60}^2-(\color{blue}{48}^2+\color{red}{36}^2)}{-2(\color{blue}{24}\times\color{red}{36})}=cos(Q)\\
$$

Answer 30

you'd needed to factor it out so you isolate the cos(), thus working only with the 3 sides, becuase is all you have in the picture

Answer 31

flutter, um. My answers keep coming out wrong. :(

Answer 32

They sensored me. lol

Answer 33

I said F u c k

Answer 34

I'm almost ready to kill this question and move on

Answer 35

hehe, so, what did you get?

Answer 36

it was a crazy number

Answer 37

ok, let's start with the numerator, what does that give you?

Answer 38

When I put it in the calc it said zero

Answer 39

right, oddly enough, \(\color{green}{60}^2-(\color{blue}{48}^2+\color{red}{36}^2)=0\) :)

Answer 40

...wow

Answer 41

Ha

Answer 42

so, anything dividing 0, the quotient is well, 0, cos cos(Q)=0

Answer 43

now just use the \(cos^{-1}(0)\), to get your number, make sure your calculator is in Degree mode

Answer 44

Wow.

Answer 45

\(
cos^{-1}[cos(Q)]=cos^{-1}[0] \implies Q = ?
\)

Answer 46

90?

Answer 47

yes, now, if you take a peek at the picture, Q looks like a right angle :)

Answer 48

Oh, you're right! Ah, thank you

Answer 49

the picture is just slanted some and vision perception can get off track sometimes

Answer 50

Ah, thanks.

Answer 51

One last question!