Question

A passenger in an airplane at an altitude of 10 kilometers sees two towns directly to the east of the plane. The angles of depression to the towns are 𝛽 = 21° and 𝜃 = 59° (see figure). How far apart are the towns? (Round your answer to one decimal place.)

 km

Answer 1

Right triangle $SBC$

$SC=10\ km, \angle SBC=\beta =21\degree$

$$BC=\dfrac{SC}{\tan \beta}=\dfrac{10}{\tan 21\degree } km$$

Right triangle $SAC$

$SC=10\ km, \angle SAC=\theta =59\degree$

$$AC=\dfrac{SC}{\tan \theta}=\dfrac{10}{\tan 59\degree } km$$

Then

$$AB=BC-AC=\dfrac{10}{\tan 21\degree } km-\dfrac{10}{\tan 59\degree } km$$

$$\approx20.0\ km$$

The towns are 20.0 km apart.