The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building
Solution:
Let the height of the tower be AB and the height of the building be CD.
The angle of elevation of the top of building D from the foot of tower B is 30° and the angle of elevation of the top of tower A from the foot of building C is 60°.
Distance between the foot of the tower and the building is BC.
Trigonometric ratio involving sides AB, CD, BC and angles ∠B and ∠C is tan θ.
In ΔABC,
tan 60° = AB/BC
√3 = 50/BC
BC = 50/√3 ....(i)
In ΔBCD,
tan 30° = CD / BC
1/√3 = CD / BC
1/√3 = CD / 50/√3 [from (i)]
CD = 1/√3 × 50/√3
CD = 50/3
Height of the building CD = 50/3 m.
☛ Check: NCERT Solutions for Class 10 Maths Chapter 9
Video Solution:
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building
Maths NCERT Solutions Class 10 Chapter 9 Exercise 9.1 Question 9
Summary:
If the angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60° and if the tower is 50 m high, then the height of the building is 50/3 m.
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