Question:
The angle of elevation of a tower PQ from a point A on the ground is observed as $30^\circ$. After walking 12 m towards the foot of the tower, the angle of tower is observed as $60^\circ$. Then the height of the tower in meter is
The angle of elevation of a tower PQ from a point A on the ground is observed as $30^\circ$. After walking 12 m towards the foot of the tower, the angle of tower is observed as $60^\circ$. Then the height of the tower in meter is
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In height and distance problems, first express the height from one triangle and substitute into the second triangle.
Updated On: Jun 15, 2026
- 18
- 9
- $6\sqrt3$
- $3\sqrt3$
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The Correct Option is C
Solution and Explanation
Concept:
Use
\[
\tan\theta=\frac{\text{height}}{\text{distance from tower}}
\]
for the two observation points.
Step 1: Assume the distance from the nearer point to the tower is $x$ m.
Then the distance from the original point A is \[ x+12 \] Let the height of the tower be \(h\). From the second position, \[ \tan60^\circ=\frac{h}{x} \] \[ h=x\sqrt3 \]
Step 2: Use the first observation.
\[ \tan30^\circ=\frac{h}{x+12} \] \[ \frac1{\sqrt3} = \frac{x\sqrt3}{x+12} \] \[ x+12=3x \] \[ 2x=12 \] \[ x=6 \]
Step 3: Calculate the height.
\[ h=x\sqrt3 \] \[ h=6\sqrt3 \]
Step 1: Assume the distance from the nearer point to the tower is $x$ m.
Then the distance from the original point A is \[ x+12 \] Let the height of the tower be \(h\). From the second position, \[ \tan60^\circ=\frac{h}{x} \] \[ h=x\sqrt3 \]
Step 2: Use the first observation.
\[ \tan30^\circ=\frac{h}{x+12} \] \[ \frac1{\sqrt3} = \frac{x\sqrt3}{x+12} \] \[ x+12=3x \] \[ 2x=12 \] \[ x=6 \]
Step 3: Calculate the height.
\[ h=x\sqrt3 \] \[ h=6\sqrt3 \]
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