# Math (last question I swear!)

Leroy is 6.5 ft tall and standing on a bridge. He looks down with an angle of depression of 31 degrees towards a boat traveling east. He then turns around and looks down with an angle of depression of 24 degrees towards a boat traveling west. If the boats are 432 ft apart, how far away from Leroy is the second boat? How do I set up this problem and explain please? Thank you!

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1. If Leroy's eyes are h feet above the water, then we can make the diagram as follows:

T is the top of Leroy's head, x feet above B, directly below him on the water.

E is the boat eastward
W is the boat westward.

Now we have

x/BW = tan 24°
x/BE = tan 31°
eliminating x, we have

BW tan 24° = BE tan 31°
But, BW+BE = 432, so
BW tan 24° = (432-BW)tan 31°
BW = 432tan31°/(tan24°+tan31°) = 248.14

The distance wanted is TW, given by

BW/TW = cos 24°
TW = BW/cos24° = 248.14/0.9135 = 271.62

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posted by Steve

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