Question

From the first floor of Qutab minar , which is at a height of 25m from the ground level , a man observes the top of a building at an angle of elevation of 30 degree and the angle of depression of the base of the building to be 60degree. Calculate the height of building

Answer 1

Let AB be the Qutub Minar and DC be the building
In right ΔABC,
AB/BC=tan 60°
25/BC=√3 ⇒25=√3 BC⇒BC=25/√3
In right ΔDBC,
DC/BC=tan 30°
DC/(25/√3)=1/√3⇒√3 DC=25/√3⇒DC=(25/√3)/√3=25/3=8.33 M
∴Height of the building=8.33 m

Answer 2

Answer:

33.33m

Step-by-step explanation:

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.