SOLUTION: a person can see the top of a building at an angle of 65 degrees the person is standing 50 ft away from the building and has an eye level of 5 ft how tall is the building to the ne
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Question 862758: a person can see the top of a building at an angle of 65 degrees the person is standing 50 ft away from the building and has an eye level of 5 ft how tall is the building to the nearest tenth of a foot Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! a person can see the top of a building at an angle of 65 degrees the person is standing 50 ft away from the building and has an eye level of 5 ft how tall is the building to the nearest tenth of a foot.
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let x= distance from eye level to top of bldg.
x/50=tan 65˚
x=tan 65˚*50
Height of building=x+5=tan 65˚*50+5=112.2 ft
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