# In triangle ABC, a=9, c=5, and angle B = 120 degrees. What is the measure of angle A to the nearest degree?

Nov 1, 2017

#### Explanation:

First, calculate the edge $b$ with the law of cosine.

${b}^{2} = {a}^{2} + {c}^{2} - 2 a c \cos B$
$= {9}^{2} + {5}^{2} - 2 \cdot 5 \cdot 9 \cdot \left(- \frac{1}{2}\right) = 151$
$b = \sqrt{151}$

Then, calculate the angle $A$ with the law of sine.
$\frac{a}{\sin A} = \frac{b}{\sin B}$
$a \sin B = b \sin A$
$\sin A = \frac{a \sin B}{b} = \frac{9 \cdot \frac{\sqrt{3}}{2}}{\sqrt{151}}$
$= \frac{9 \sqrt{453}}{302}$
≒0.6343

Note that $A$ must be smaller than $180 - 120 = 60$ degree.
Thus, angle $A$ is
A=arcsin(0.6343)≒39.37 deg.