Given an arbitrary non-right triangle, we can drop an altitude, which we temporarily label h,
to create two right triangles.
From right triangle ABD
x=tanαh
From right triangle BCD
y=tanβh
Then
AC=x+y=tanαh+tanβh=h(tanαtanβtanα+tanβ)
Solve for
h=tanα+tanβtanαtanβ⋅AC
h=sinαcosβ+sinβcosαsinαsinβ⋅AC
h=sin(α+β)sinαsinβ⋅AC
Given α=18°,β=32°,AC=32 miles.
Substitute
h=sin(18°+32°)sin18°sin32°⋅30 mi≈6.413 mi The altitude of the aircraft is 6.413 miles.
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