Question

From The Top Of A Building 30M High The Angles Of Depression Of The Top And The Foot Of The Tower Observed To Be 30 Degree And 60 Degree Respectively Find The Height Of The Tower.​

Answer 1

Correct Question :-

From the top of a building 30m high, the angles of depression of the top and the foot of the tower observed to be 30° and 60° respectively. Find the height of the tower.

Answer:

The height of the tower is 20m.

Step-by-step Explanation

To Find -

• The height of the tower.

★ Solution :-

Given that,

• The top of a building is 30m high.
• The angles of depression of the top and the foot of the tower observed to be 30° and 60° respectively.

Alternate angles: ∠MPN = 60° and ∠MOE = 30°.

Diagram: Refer the attachment.

Assumption

Let us consider MN as the building and OP as the tower.

In right ∆MNP,

• tan 60° = MN/NP

[tan 60° = MN/NP ∵ tan 60° = Opposite/Adjacent].

Therefore,

⟹ tan 60° = MN/NP

⟹ √3 = MN/NP

⟹ √3 = 30/NP

⟹ NP = 30/√3

⟹ NP = 10√3

Hence, OE = NP = 10√3.

Now, In right ∆MEO,

NE = OP, MN = 30m (given), assuming OP as "x" metres.

ME = MN - NE,

⟹ ME = MN - NE = 30 - OP = 30 - x.

Now, tan 30° = ME/OE.

We know, tan 30° = 1/√3

⟹ tan 30° = ME/OE

⟹ 1/√3 = (30 - x)/10√3

⟹ 1(10√3) = √3(30 - x)

⟹ 10√3 = √3(30 - x)

Cancelling √3 from both sides,

⟹ 10 = 30 - x

⟹ 10 - 30 = - x

⟹ - 20 = - x

Cancelling negative signs from both sides,

⟹ 20 = x

Hence, The height of the tower is 20m.