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

Step 1: Understanding the Concept:
This is an application of trigonometry involving angles of depression. Angles of depression from the top of the taller building are equal to the angles of elevation from the top and bottom of the shorter building due to alternate interior angles.

Step 2: Setting up the Triangles:
Let the height of the multi-storeyed building be \( H = (h + 6) \) m and the distance between buildings be \( x \) m.
In the small right triangle (to the top of the 6 m building):
\( \tan 30^\circ = \frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{x} \Rightarrow x = h\sqrt{3} \)   --- (Eq. 1)

In the large right triangle (to the bottom of the 6 m building):
\( \tan 60^\circ = \frac{h + 6}{x} \Rightarrow \sqrt{3} = \frac{h + 6}{x} \Rightarrow x = \frac{h + 6}{\sqrt{3}} \)   --- (Eq. 2)

Step 3: Solving for h and H:
Equating \( x \) from Eq. 1 and Eq. 2:
\( h\sqrt{3} = \frac{h + 6}{\sqrt{3}} \)
\( 3h = h + 6 \Rightarrow 2h = 6 \Rightarrow h = 3 \) m.
Height of multi-storeyed building \( H = 3 + 6 = 9 \) m.
Distance \( x = 3\sqrt{3} \) m \( \approx 3 \times 1.732 = 5.196 \) m.

Step 4: Final Answer:
The height of the multi-storeyed building is 9 m and the distance between them is \( 3\sqrt{3} \) m.

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Solution (OR Question):

Step 1: Understanding the Concept:
We have two poles of equal height \( h \) on either side of a 60 m road. A point on the road creates two right-angled triangles with different angles of elevation.

Step 2: Setting up Equations:
Let height of poles be \( h \). Let the distance of the point from the first pole be \( y \). Then its distance from the second pole is \( (60 - y) \).

For the \( 60^\circ \) elevation:
\( \tan 60^\circ = \frac{h}{y} \Rightarrow \sqrt{3} = \frac{h}{y} \Rightarrow h = y\sqrt{3} \)   --- (Eq. 1)

For the \( 30^\circ \) elevation:
\( \tan 30^\circ = \frac{h}{60 - y} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{60 - y} \Rightarrow h = \frac{60 - y}{\sqrt{3}} \)   --- (Eq. 2)

Step 3: Solving for y and h:
\( y\sqrt{3} = \frac{60 - y}{\sqrt{3}} \)
\( 3y = 60 - y \Rightarrow 4y = 60 \Rightarrow y = 15 \) m.
\( h = 15\sqrt{3} \approx 25.98 \) m.
Distances from poles: 15 m and \( 60 - 15 = 45 \) m.

Step 4: Final Answer:
The height of the poles is \( 15\sqrt{3} \) m and the distances from the poles are 15 m and 45 m.