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 east of the plane the angles of depression to the towns of 28° and 55°. How fa class=

Answer :

Answer:

The two towns are 62.57km apart.

Explanation:

A diagram representing the problem is redrawn and attached below:

From trigonometric ratios:

[tex]\begin{gathered} \tan 55\degree=\frac{53}{OA} \\ OA\times\tan 55\degree=53 \\ OA=\frac{53}{\tan 55\degree} \\ OA=37.11\operatorname{km} \end{gathered}[/tex]

Similarly:

[tex]\begin{gathered} \tan 28\degree=\frac{53}{OB} \\ OB\times\tan 28\degree=53 \\ OB=\frac{53}{\tan 28\degree} \\ OB=99.68\operatorname{km} \end{gathered}[/tex]

Therefore, the distance between the two towns is:

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

The two towns are 62.57km apart.

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