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A passenger in an airplane flying at an altitude of 37,000 feet sees two towns directly to the west of the airplane. The angles of depression to the towns are 32° and 76°. How far apart are the towns?
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A passenger in an airplane flying at an altitude of 37000 feet sees two towns directly to the west of the airplane The angles of depression to the towns are 32 class=

Respuesta :

calculista

Answer:

[tex]49,987.24\ ft[/tex]

Step-by-step explanation:

Let

x-----> distance from the two towns

we know that

In the right triangle ABC

[tex]tan(32\°)=37,000/(x+y)\\x+y=37,000/tan(32\°)[/tex]

In the right triangle ABD

[tex]tan(76\°)=37,000/(y)\\y=37,000/tan(76\°)[/tex]

see the attached figure to better understand the problem

Remember that

(x+y)-y=x

so

[tex]x=\frac{37,000}{tan(32\°)} -\frac{37,000}{tan(76\°)}[/tex]

[tex]x=49,987.24\ ft[/tex]

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