Question

In △ABC, m∠A=15°, a=9, and b=10. Find c to the nearest tenth.

Answer 1

In △ABC, m∠A=15°, a=9, and b=10

We apply sine law

$\frac{ sin(a)}{b} = \frac{sin(b)}{b} = \frac{sin(c)}{c}$

$\frac{ sin(a)}{b} = \frac{sin(b)}{b}$

$\frac{ sin(15)}{9} = \frac{sin(b)}{10}$

$\frac{sin 15}{9} * 10 = sin(b)$

0.287576716 = sin(b)

b= 16.7129 degree

angle A= 15 degree, angle B = 16.71 degree

So angle C = 148.29

$\frac{ sin(a)}{b} = \frac{sin(c)}{c}$

$\frac{ sin(15)}{9} = \frac{sin(148.29)}{c}$

c= $\frac{sin(148.29)}{sin(15)} *9$

c= 18.28