Question

Two radar towers located 30 miles apart each detect an aircraft flying between them.

The angle of elevation measured by the first station is 32°. The angle measured by the second station is 18°.

Find the altitude of the aircraft.

   

Answer 1

Given an arbitrary non-right triangle, we can drop an altitude, which we temporarily label $h,$

to create two right triangles.

From right triangle $ABD$

$$x=\dfrac{h}{\tan \alpha}$$

From right triangle $BCD$

$$y=\dfrac{h}{\tan \beta}$$

Then

$$AC=x+y=\dfrac{h}{\tan \alpha}+\dfrac{h}{\tan \beta}=h(\dfrac{\tan \alpha+\tan \beta}{\tan \alpha\tan \beta})$$

Solve for

$$h=\dfrac{\tan \alpha\tan \beta}{\tan \alpha+\tan \beta}\cdot AC$$

$$h=\dfrac{\sin \alpha\sin \beta}{\sin \alpha\cos \beta+\sin \beta\cos \alpha}\cdot AC$$

$$h=\dfrac{\sin \alpha\sin \beta}{\sin (\alpha+\beta)}\cdot AC$$

Given $\alpha=18\degree, \beta=32\degree, AC=32 \ miles.$

Substitute

$$h=\dfrac{\sin18\degree\sin 32\degree}{\sin (18\degree+32\degree)}\cdot 30\ mi\approx6.413 \ mi$$

The altitude of the aircraft is 6.413 miles.