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Q) 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, then find the height of the building.

Ans:

Step 1: Diagram for this question:

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, then find the height of the building.

Let AB be the tower of 50 m height and PQ be the building of H m.

The distance between the building and the tower be D m, and angle of elevations be as shown in the image above.

Step 2: Find the distance D:

In Δ ABQ, tan 60 = \frac{AB}{BQ}

\sqrt 3 = \frac{50}{D}

∴  D = \frac{50}{\sqrt 3} 

Step 3: Find the height H:

Now in Δ PQB, tan 30 = \frac{PQ}{QB}

\frac{1}{\sqrt 3} = \frac{H}{D}

\therefore H = \frac{D}{\sqrt 3} = \frac{50}{\sqrt 3 \times \sqrt 3}

or H = \frac{50}{3} m

Therefore, the height of the building is \frac{50}{3} m.

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