Precalculus (6th Edition) Blitzer

The missing value are: $A=39{}^\circ,B=78{}^\circ \text{ and }C=63{}^\circ$.
For any triangle, The law of sines states that: $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\operatorname{sinC}}$ The law of cosines states that: \begin{align} & {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2\cdot b\cdot c\cdot \cos A \\ & {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\ & {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2\cdot a\cdot b\cdot \cos C \\ \end{align} Use the law of cosines to obtain: \begin{align} & {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2\cdot a\cdot c\cdot \cos B \\ & {{40.2}^{2}}={{26.1}^{2}}+{{36.5}^{2}}-2\cdot \left( 26.1 \right)\cdot \left( 36.5 \right)\cdot \cos B \\ & \cos B=\frac{\left( {{26.1}^{2}}+{{36.5}^{2}} \right)-{{40.2}^{2}}}{2\cdot \left( 26.1 \right)\cdot \left( 36.5 \right)} \\ & \cos B\approx 0.20 \end{align} We will simplify it further to get the measure of angle B. \begin{align} & \cos B\approx 0.20 \\ & B\approx {{\cos }^{-1}}\left( 0.20 \right) \\ & B\approx 78{}^\circ \end{align} Using the law of sines we get, \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{26.1}{\sin A}=\frac{40.2}{\sin 78{}^\circ } \\ & 40.2\left( \sin A \right)=26.1\left( \sin 78{}^\circ \right) \\ & \sin A=\frac{26.1\left( \sin 78{}^\circ \right)}{40.2} \end{align} We will simplify it further to obtain the measure of angle A. \begin{align} & \sin A=0.63 \\ & A={{\sin }^{-1}}\left( 0.63 \right) \\ & A\approx 39{}^\circ \end{align} Using the angle sum property we will obtain the measure of angle C as: \begin{align} & A+B+C=180{}^\circ \\ & 39{}^\circ +78{}^\circ +C=180{}^\circ \\ & C=180{}^\circ -117{}^\circ \\ & C=63{}^\circ \end{align} So, $A=39{}^\circ,B=78{}^\circ \text{ and }C=63{}^\circ$. Hence, $A=39{}^\circ,B=78{}^\circ \text{ and }C=63{}^\circ$.