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.
Show complete solution and explaoin the answer.
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)