Answer to Question #206421 in Trigonometry for Oula

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.