## Precalculus (6th Edition) Blitzer

The solution is $A=26{}^\circ,C=44{}^\circ \ \text{ and }\ b=21.6$
The given angles and sides of the triangle are: $B=110{}^\circ,a=10,c=16$ Using the law of cosines, we will find the side b of the triangle. That is, \begin{align} & {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cos B \\ & {{b}^{2}}={{10}^{2}}+{{16}^{2}}-2\left( 10 \right)\left( 16 \right)\cos 110{}^\circ \\ & {{b}^{2}}=465.4464 \\ & b=21.6 \end{align} Now, by using the law of sines we will find the angle A of the triangle. That is, \begin{align} & \frac{\sin A}{a}=\frac{\operatorname{sinB}}{b} \\ & \frac{\sin A}{10}=\frac{\sin 110{}^\circ }{21.6} \end{align} This implies that, \begin{align} & \sin A=10\times \frac{\sin 110{}^\circ }{21.6} \\ & =0.4350 \end{align} Therefore, $A=26{}^\circ$ Now, the third angle will be found as below: \begin{align} & C=180{}^\circ -A-B \\ & =180{}^\circ -26{}^\circ -110{}^\circ \\ & C=44{}^\circ \end{align} Therefore, the solution is: $A=26{}^\circ,C=44{}^\circ \ \text{ and }\ b=21.6$