In Triangle ABC, a=15cm, c=9cm, and angle C= 35 degrees. Find b and angle b. (Recall sin theta= sin (180 degrees-theta). I'm not sure how to approach this.

Answer 1

you have C and c and a, so you can find A using

sinA/a = sinC/c

Now, having A and C, you know that A+B+C=180, so you can easily find B.

Then, use the law of sines again to get b:

b/sinB = c/sinC = a/sinA

Answer 2

sin A/15 = sin 35/9 = .0637

sin A = .956

so A = 72.9 degrees or 180-72.9 = 107.1
then A + B + C = 180
72.9 + B + 35 = 180
B = 72.1 degrees (One answer)
sin B/b = .0637
so b = 14.94
or
A = 107.1
then B = 180 - 35 -107.1 = 37.9 degrees (alternate answer)

Answer 3

Thanks! I should have mentioned this in the question, but the answer sheet asks for two values of angle A rather than 1.

Answer 4

Oh nvm thanks Damon.

Answer 5

And steve sorry I didn't read the whole answer.