Question:medium

The angles of elevation of the top of a tower from two points at a distance of 5 meters and 20 meters along the same straight line from the base of the tower, are complementary. Find the height of the tower.

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There is a direct formula for this specific problem: If the angles of elevation of the top of a tower from two points at distances 'a' and 'b' from its base are complementary, then the height of the tower 'h' is given by \( h = \sqrt{ab} \). Here, \(h = \sqrt{5 \times 20} = \sqrt{100} = 10\) m.
Updated On: Mar 29, 2026
  • 10 m
  • 15 m
  • \(10\sqrt{3}\) m
  • 20 m
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The Correct Option is A

Solution and Explanation

Step 1: Conceptualization:
This is a trigonometry problem involving angles of elevation and complementary angles (summing to 90 degrees). The tangent function is used to relate the tower's height to distances and angles.
Step 2: Method Identification:
Let 'h' be the tower height.
Let the observation points be at distances 'a' (5m) and 'b' (20m) from the tower base.
Let the angles of elevation be \( \theta \) and \( 90^{\circ} - \theta \). The height of the tower can be calculated using \(h = \sqrt{ab}\).
Step 3: Derivation:
Let the tower height be h. The observation points are at 5 m and 20 m from the base.
For the point at 20 m, the angle of elevation is \( \theta \).
Therefore, \( \tan(\theta) = \frac{h}{20} \) -- (1)
For the point at 5 m, the angle of elevation is \( 90^{\circ} - \theta \).
Therefore, \( \tan(90^{\circ} - \theta) = \frac{h}{5} \) -- (2)
Using the identity \( \tan(90^{\circ} - \theta) = \cot(\theta) \), equation (2) becomes:
\( \cot(\theta) = \frac{h}{5} \)
Since \( \cot(\theta) = \frac{1}{\tan(\theta)} \), multiplying the tangent expressions from (1) and (2):
\( \tan(\theta) \times \cot(\theta) = \left(\frac{h}{20}\right) \times \left(\frac{h}{5}\right) \)
Given \( \tan(\theta) \times \cot(\theta) = 1 \):
\[ 1 = \frac{h^2}{100} \]
\[ h^2 = 100 \]
\[ h = \sqrt{100} = 10 \] m (Height is a positive value).
Step 4: Conclusion:
The tower's height is 10 m.
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