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DiKsa [7]
3 years ago
7

An ant looks to the top of a bulding at an angle of elevation of 32°. The ant then walks an additional 66 feet from the edge of

the building. If the new angle of elevation from the ant to the top of the building is 22°, find the height of the building. Round to the nearest tenth
Mathematics
1 answer:
Keith_Richards [23]3 years ago
8 0
Let the height of the building be x. Let the initial distance of the ant from the building be y, then
tan 32 = x/y
y = x/tan 32 . . . . . . . . (1)
tan 22 = x/(y + 66)
y tan 22 + 66 tan 22 = x
y = (x - 66 tan 22)/tan 22 . . . . . . . . (2)
Equating (1) and (2), we have
x/tan 32 = (x - 66 tan 22)/tan 22
x tan 22 = x tan 32 - 66 tan 22 (tan 32)
x(tan 32 - tan 22) = 66 tan 22 (tan 32)
x = (66 tan 22 (tan 32))/(tan 32 - tan 22) = 66(0.4040)(0.6249)/(0.6249 - 0.4040) = 16.6626/0.2208 = 75.4

Therefore, the height of the building is 75.4 feet.
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Figure show that triangle ABC and AD is the height of the triangle.

Since, the triangle ABC is an equilateral triangle,

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Using Pythagoras theorem,

AD^{2} +BD^{2} =AC^{2}

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Hope this helps!
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