## Precalculus (6th Edition) Blitzer

The value of $c$ to the nearest tenth is $6.2$.
Using the Law of Cosines, we get, ${{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C$ Substitute the values $C=68{}^\circ,a=5,b=6$ ; So, \begin{align} & {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\cos C \\ & {{c}^{2}}={{5}^{2}}+{{6}^{2}}-2\left( 5 \right)\left( 6 \right)\cos 68{}^\circ \\ & =61-60\cos 68{}^\circ \\ & =38.52 \end{align} Further, \begin{align} & c=\sqrt{38.52} \\ & \approx 6.2 \end{align} So, the value of $c$ is $6.2$. Therefore, the value of $c$ to the nearest tenth is $6.2$.