Question

Will Medal!

The measure of theta, in degrees, is approximately 48.8 degrees?

A. True
B. False

(Picture Below)
***My Answer: B***

Answer 1

do have any reasons why you think it's B?

Answer 2

did you use the law of Cosines?
cos A = \[\cos A = \frac{(b^2+c^2-a^2)}{2bc}\]
\[\cos B = \frac{(c^2+a^2-b^2)}{2ca} \]

Answer 3

I used the formula:
\[c ^{2} = a ^{2} + b ^{2} -2ab cosC\]

Answer 4

Couldn't you just use arccos?

Answer 5

hmmm but I want cos C to be by itself

and it's easier to grab the angle A and B first.

Answer 6

Now that I see the formulas you gave me, I think I may be doing the wrong formula.

Answer 7

@zeesbrat3 arccos is later into the formula... you have to find angle A and angle B first .. since this is all S S S (side side side)

Answer 8

Fair enough

Answer 9

c is supposed to be the longest side a.k.a hypotenuse

Answer 10

Oh ok. The laws of cosines were getting me confused for a second.

Answer 11

So we would have a = 5.3 b = 4 and c = 7?

Answer 12

I need letters first... we need to redraw the triangle with A B C and then look across for our a b c ok let's use those values

Answer 13

so that would mean ... a = 5.3 , b=4, and c = 7...
now we use Cos A to find Angle A

Answer 14

Ok

Answer 15

\[\cos A = \frac{(b^2+c^2-a^2)}{2bc} \]
you just have to plug in those values.. and simplify

Answer 16

Great! Thank you! So we would have cos A = 0.66?

Answer 17

Then in degrees it would be 37.8 degrees?

Answer 18

this is where you need to use \[\cos^{-1}(0.66) \] and then you use a calculator to find the degrees of it

Answer 19

Can I just double check this for a sec? ^_^ so fast

Answer 20

Oh ok! Yeah go ahead :)

Answer 21

\[\cos A = \frac{(4^2+7^2-5.3^2)}{2(4)(7)}\]

Answer 22

Oh nevermind I got 48.7 degrees

Answer 23

I got \[\cos \theta = .6835\]

Answer 24

\[\cos A = \frac{(16+49-28.09)}{56} \rightarrow \frac{36.91}{56} \rightarrow 0.659 \rightarrow 0.66\]

Answer 25

\[\cos^{-1}(0.66) \rightarrow 48.7\] I also got 48.7 degrees

Answer 26

My bad, I used 54 as the denominator, not 56... ignore the blonde...

Answer 27

ok cool... we got angle A = 48.7 degrees...
Now we need angle B

Answer 28

Oh whoops for the the rest of the equation haha

Answer 29

let's try and solve further.

Answer 30

Ok

Answer 31

\[\cos B = \frac{(c^2+a^2-b^2)}{2ca} \]
c = 7, b = 4, a = 5.3

Answer 32

back the equation tool gave me a hard time

Answer 33

I got around 35.1 degrees (34.95 not rounded)

Answer 34

\[\cos B = \frac{(7^2+5.3^2-4^2)}{2(7)(5.3)}\]

\[\cos B = \frac{(49+28.09-16)}{74.2}\]
\[\cos B = \frac{61.09}{74.2}\]
\[\cos B = 0.84\]
\[\cos^{-1}(0.84) =31.95 -> 32\]

huh that's odd... how did you get 35.1 degrees?

Answer 35

another approach
when given only the 3 sides use cosine rule to find largest angle opposite longest side

Answer 36

oh wait I got the division wrong
\[\frac{61.09}{74.2} = 0.823 \]

Answer 37

\[\cos^{-1}(0.823) = 34.58\]

Answer 38

anyway .. let's just assume that Angle B = 34.58
Angle A = 48.7

to find Angle C.. we use the fact that angles of a triangle must add up to 180 degrees

Angle C = 180-48.7-34.58

I will also take the 35.1 into consideration for Angle B

Angle C = 180-48.7-35.1