From the first floor of Qutab Minar, which is at a height of 25 m fro

1.	From the first floor of Qutab Minar, which is at a height of 25 m from the ground level, a manobserves the top of a building at an angle of elevation of 30Â° and the angle of depression of thebase of the building to be 60Â°. Calculate the height of the building.
Answer» Let the height of the first floor of qutub minar be AB=EC=25m And DEC (DC) be the height of the tower. BC=AE= Distance between their bases Angle DAE be the angle of elevation of the top of the building. Angle EAC be the angle of depression of the base of the building. In triangle EAC, Tan angle EAC=EC/AE => √3=25/AE => AE=25/ √3 meters In triangle ADE, Tan angle DAE=DE/AE => 1/ √3= DE/25/√3 => 3 DE=25 => DE=25/3 => DE=8.33 approx Height of the building= DE+EC = 8.33+25=33.33 meters approx. Thus, the height of the building=33.33 metres.

From the first floor of Qutab Minar, which is at a height of 25 m from the ground level, a manobserves the top of a building at an angle of elevation of 30Â° and the angle of depression of thebase of the building to be 60Â°. Calculate the height of the building.

Answer»

Let the height of the first floor of qutub minar be AB=EC=25m

And DEC (DC) be the height of the tower.

BC=AE= Distance between their bases

Angle DAE be the angle of elevation of the top of the building.

Angle EAC be the angle of depression of the base of the building.

In triangle EAC,

Tan angle EAC=EC/AE

=> √3=25/AE

=> AE=25/ √3 meters

In triangle ADE,

Tan angle DAE=DE/AE

=> 1/ √3= DE/25/√3

=> 3 DE=25

=> DE=25/3

=> DE=8.33 approx

Height of the building= DE+EC = 8.33+25=33.33 meters approx.

Thus, the height of the building=33.33 metres.