## Precalculus (6th Edition) Blitzer

The two airplanes would be $861\text{ miles}$ apart.
The distance traveled by the first airplane in $2$ hours is: $325\times 2=650\text{ miles}$ The distance covered by the second airplane in 2 hours is: $300\times 2=600\text{ miles}$ We can illustrate the given situation by the figure given below, where the planes started from point A and the distance between them after 2 hours would be BC or a. At first, we will compute the measure of angle BAC using the straight angle property: \begin{align} & 66.5{}^\circ +\angle BAC+26.5{}^\circ =180{}^\circ \\ & \angle BAC=180{}^\circ -66.5{}^\circ -26.5{}^\circ \\ & =180-93{}^\circ \\ & =87{}^\circ \end{align} Now consider triangle ABC, Here, $A=87{}^\circ,b=600,c=650$. Using the law of cosines we will evaluate the side BC as: \begin{align} & B{{C}^{2}}=A{{C}^{2}}+A{{B}^{2}}-2\cdot AC\cdot AB\cdot \cos A \\ & B{{C}^{2}}={{600}^{2}}+{{650}^{2}}-2\left( 600 \right)\left( 650 \right)\cos 87{}^\circ \\ & B{{C}^{2}}=741,678 \\ & BC\approx 861 \end{align} So, the distance between the airplanes after 2 hours will be $861$ miles.