Question

The angles of depression of the top and the bottom of a 6 m high building from the top of a multi-storeyed building are 30° and 60° respectively. Find the height of the multi-storeyed building and the distance between the two buildings.

OR

Two poles of equal heights are standing opposite each other on either side of the road, which is 60 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 30° and 60°, respectively. Find the height of the poles and the distances of the point from the poles.

Hint:

In these problems, always draw a clear diagram first. The "angle of depression" from the top is numerically equal to the "angle of elevation" from the bottom.

Answer 1

Solution 1

Step 1: Let the required quantities be variables.
Let the height of the multi-storeyed building be \(h\) m and the distance between the two buildings be \(x\) m.
The other building has height 6 m.

Step 2: Use the angle of depression to the bottom of the building.
The angle of depression from the top of the multi-storeyed building to the bottom of the 6 m building is 60°.
\[ \tan 60^\circ = \frac{\text{height of multi-storeyed building}}{\text{distance}} \] \[ \sqrt{3} = \frac{h}{x} \] \[ h = x\sqrt{3} \]

Step 3: Use the angle of depression to the top of the building.
The difference in height between the two tops is \(h - 6\). The angle of depression is 30°.
\[ \tan 30^\circ = \frac{h - 6}{x} \] \[ \frac{1}{\sqrt{3}} = \frac{h - 6}{x} \] \[ h - 6 = \frac{x}{\sqrt{3}} \]

Step 4: Substitute the value of \(h\).
\[ x\sqrt{3} - 6 = \frac{x}{\sqrt{3}} \] Multiply by \(\sqrt{3}\): \[ 3x - 6\sqrt{3} = x \] \[ 2x = 6\sqrt{3} \] \[ x = 3\sqrt{3} \]

Step 5: Find the height of the multi-storeyed building.
\[ h = x\sqrt{3} \] \[ h = 3\sqrt{3} \times \sqrt{3} \] \[ h = 9 \]

Final Answer:
Height of the multi-storeyed building = 9 m
Distance between the buildings = 3√3 m.

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Solution 2

Step 1: Assume the distances.
Let the distance of the point from the first pole be \(x\) m and from the second pole be \(60 - x\) m.
Let the height of each pole be \(h\) m.

Step 2: Use the angle of elevation 30°.
\[ \tan 30^\circ = \frac{h}{x} \] \[ \frac{1}{\sqrt{3}} = \frac{h}{x} \] \[ h = \frac{x}{\sqrt{3}} \]

Step 3: Use the angle of elevation 60°.
\[ \tan 60^\circ = \frac{h}{60 - x} \] \[ \sqrt{3} = \frac{h}{60 - x} \] \[ h = \sqrt{3}(60 - x) \]

Step 4: Equate both expressions for height.
\[ \frac{x}{\sqrt{3}} = \sqrt{3}(60 - x) \] Multiply by \(\sqrt{3}\): \[ x = 3(60 - x) \] \[ x = 180 - 3x \] \[ 4x = 180 \] \[ x = 45 \]

Step 5: Find the height of the poles.
\[ h = \frac{45}{\sqrt{3}} \] \[ h = 15\sqrt{3} \]

Final Answer:
Height of each pole = 15√3 m
Distance from the first pole = 45 m
Distance from the second pole = 15 m.