Question #206421

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.

   


1
Expert's answer
2021-06-15T08:38:30-0400

Given an arbitrary non-right triangle, we can drop an altitude, which we temporarily label h,h,

to create two right triangles.




From right triangle ABDABD


x=htanαx=\dfrac{h}{\tan \alpha}

From right triangle BCDBCD

y=htanβy=\dfrac{h}{\tan \beta}


Then


AC=x+y=htanα+htanβ=h(tanα+tanβtanαtanβ)AC=x+y=\dfrac{h}{\tan \alpha}+\dfrac{h}{\tan \beta}=h(\dfrac{\tan \alpha+\tan \beta}{\tan \alpha\tan \beta})


Solve for


h=tanαtanβtanα+tanβACh=\dfrac{\tan \alpha\tan \beta}{\tan \alpha+\tan \beta}\cdot AC

h=sinαsinβsinαcosβ+sinβcosαACh=\dfrac{\sin \alpha\sin \beta}{\sin \alpha\cos \beta+\sin \beta\cos \alpha}\cdot AC

h=sinαsinβsin(α+β)ACh=\dfrac{\sin \alpha\sin \beta}{\sin (\alpha+\beta)}\cdot AC

Given α=18°,β=32°,AC=32 miles.\alpha=18\degree, \beta=32\degree, AC=32 \ miles.

Substitute


h=sin18°sin32°sin(18°+32°)30 mi6.413 mih=\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.



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