Question

Trig question: needs checking.
Diagrams and calculations below

Answer 1

length from a to b is 37.66km? and length from b to c is 2km?

Answer 2

What I did (all trig functions are in degrees):\[b^{2}=a^{2}+c^{2}-2ac\cos B\]\[=2^{2}+37.66^{2}-2(2)(37.66)\cos 132\]\[\approx1321.48\]

Answer 3

@TylerD Yes

Answer 4

have you tried law of sines?

Answer 5

I used the Cosines Law.
I'm wondering if I did it right...

Answer 6

looks right

Answer 7

but did u take the square root after

Answer 8

im getting 39.0265km for b

Answer 9

wait a sec

Answer 10

Okay thanks.
Since they want angle A, can I use the value of "segment b" that I found to do the Sine Law? :o
(I lagged out sorry)

yeah sorry I forgot to square root, you're right...
But I got 36.35km.

Answer 11

now that you have an angle and the side, you can use sine law but

Answer 12

I calculated sqrt and got 36.___ both ways o_o; did I do something wrong

Answer 13

2^2+37.66^2=1422.2756

1422.2756-(2*2*37.66cos(132)) = 1523.073435

sqrt(1523.073435)=39.02657344

Answer 14

O_O I didn't get that at all...

Answer 15

I re-calculated and got 1550.0256

Answer 16

sin(132)/39.03 = sin(A)/2

2(sin(132))/39.03=sin(A)

sin^-1(2(sin(132))/39.03)=A

Answer 17

OH I did addition on accident! *facepalms*

Answer 18

Okay, so I got 1523.07 which square roots to 39.02...

Answer 19

right now use law of sines.

Answer 20

and then I should be good for angle A?

Answer 21

and then solve for theta with a sin inverse

Answer 22

ya

Answer 23

Okay

Answer 24

Thank you ! :)