An observer 30 m away from the base of a building finds the angle of elevation of its top to be 30°. He then walks towards the building until the angle of elevation becomes 45°. How far (in m) did he walk?

Question

Accepted Answer

30 - 10 √3. Height h = 30 tan 30 = 30√3 = 10 √3. For 45°: tan 45 = h/d2 => d2 = h = 10 √3. Walked = 30 - 10 √3 m.

An observer 30 m away from the base of a building finds the angle of elevation of its top to be 30°.
He then walks towards the building until the angle of elevation becomes 45°. How far (in m) did he walk?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Heights and Distances practice

An observer 30 m away from the base of a building finds the angle of elevation of its top to be 30°.He then walks towards the building until the angle of elevation becomes 45°. How far (in m) did he walk?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Heights and Distances practice

An observer 30 m away from the base of a building finds the angle of elevation of its top to be 30°.
He then walks towards the building until the angle of elevation becomes 45°. How far (in m) did he walk?