## Precalculus (6th Edition) Blitzer

The required solution is $\text{698}\ \text{miles}$
The radius of the earth $r$ is $4000\ \text{miles}$. The angle $\theta$ at the center between A and B is $10{}^\circ$. So, convert $\theta$ into radians: \begin{align} & \theta =10{}^\circ \left( \frac{\pi }{180{}^\circ } \right) \\ & =\frac{\pi }{18} \end{align} And the distance $s$ between A and B is given by $s=r\theta$ Put $4000\ \text{miles}$ for $r$ and $\frac{\pi }{18}$ for $\theta$: \begin{align} & s=\left( 4000\ \text{miles} \right)\left( \frac{\pi }{18} \right) \\ & =\frac{2000\pi }{9}\ \text{miles} \end{align} Put $\pi =3.14159$: \begin{align} & s=\frac{2000\left( 3.14159 \right)}{9}\ \text{miles} \\ & \text{=698}\text{.13}\approx \text{698}\ \text{miles} \end{align}