A 50-foot building is 100 feet away from another building. If the angle of elevation from the 50-foot building to the... (continued)
...5th floor of the other building is 10° and the angle of depression from the 50-foot building to the 2nd floor of the other building is 8°, find the distance of the vertical distance of the 2nd floor and the 3rd floor of the other building.
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The vertical distance from the top of the 50 foot building down to the 2nd floor of the opposite building is given by:
`tan 8^(circ)=d/100=>d~~14.05` feet. (`d=100 tan 8^(circ)` )
The vertical distance from the top of the 50 foot building to the fifth floor of the building opposite can be found by:
`tan10^(circ)=d'/100 => d'~~17.63` feet. (`d'=100 tan10^(circ)` )
Thus the distance from the 2nd floor to the 5th floor is `d+d'~~31.68` feet.
If we assume that each floor of the building is of the same height, then the distance from the 2nd floor to the 3rd floor is approximately `31.68/3=10.56` feet.
(I'm not at all sure this is an appropriate assumption, as the first floor is `50-14.05=35.95` feet -- this seems odd, but it is the only reasonable way to answer the question as stated)
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