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a passenger in an airplane at an altitude of 53 km sees two towns directly to the east of the plane the angles of depression to the towns of 28° and 55°. How far apart are the towns? round Answer to two decimal places.

a passenger in an airplane at an altitude of 53 km sees two towns directly to the-example-1
User Andre
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1 Answer

3 votes

Answer:

The two towns are 62.57km apart.

Step-by-step explanation:

A diagram representing the problem is redrawn and attached below:

From trigonometric ratios:


\begin{gathered} \tan 55\degree=(53)/(OA) \\ OA*\tan 55\degree=53 \\ OA=(53)/(\tan 55\degree) \\ OA=37.11\operatorname{km} \end{gathered}

Similarly:


\begin{gathered} \tan 28\degree=(53)/(OB) \\ OB*\tan 28\degree=53 \\ OB=(53)/(\tan 28\degree) \\ OB=99.68\operatorname{km} \end{gathered}

Therefore, the distance between the two towns is:


\begin{gathered} AB=OB-OA \\ =99.68-37.11 \\ =62.57\operatorname{km} \end{gathered}

The two towns are 62.57km apart.

a passenger in an airplane at an altitude of 53 km sees two towns directly to the-example-1
User Akronix
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