# How do you solve a triangle given a=10 b=8 B=130 degrees?

Nov 9, 2015

#### Explanation:

The Law of Sines states

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

We are given $a$ $b$ and $B$ so we can set them equal to each other and solve for Sin of A

$\frac{10}{\sin A}$ = $\frac{8}{\sin 130}$

manipulating that you get $\frac{10 \sin 130}{8}$ = $0.95755$

Taking the inverse ${\sin}^{-} 1 \left(0.95755\right)$ = $73.13$

This is the angle A. If you have 2 angles of a triangle you really have 3 angles. Just take the sum of the two known angles and subtract them from 180.

$\left(130 + 73.13\right) - 180 = 23.13$

Now that you have all three angles you can use the cosine equation

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos C$

${c}^{2} = {8}^{2} + {10}^{2} - \left(10 \cdot 8\right) \cos 23.13$

${c}^{2} = 16.86$

$\sqrt{{c}^{2}} = \sqrt{16.86}$

$c = 4.1$

I hope that helps.