Answer
The two airplanes would be $861\text{ miles}$ apart.
Work Step by Step
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.
