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The tower is 650 meters high. Suppose a building is erected such that the base of the building is on the same plane as the base of the​ tower, the angle of elevation from the top of the building to the top of the tower is 80.38° and the angle of depression from the top of the building to the foot of the tower is 65.05​°. How high would the building have to​ be?

User RWIL
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1 Answer

5 votes

Answer:


173.57\ m

Explanation:

Let

h -----> the height of the building

x -----> the horizontal distance between the building and the tower

we know that


tan(80.38\°)=(650-h)/x

solve for x


x=(650-h)/tan(80.38\°) -----> equation A


tan(65.05\°)=h/x

solve for x


x=h/tan(65.05\°) ------> equation B

equate equation A and equation B and solve for h


h/tan(65.05\°)=(650-h)/tan(80.38\°)


tan(80.38\°)h=(650-h)tan(65.05\°)


tan(80.38\°)h=(650)tan(65.05\°)-(h)tan(65.05\°)


h[tan(80.38\°)+tan(65.05\°)]=(650)tan(65.05\°)


h=(650)tan(65.05\°)/[tan(80.38\°)+tan(65.05\°)]


h=173.57\ m

User Kreuzberg
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