Question:

From a point on a bridge across a river, the angles of depressions of the banks on opposite sides of the river are \(30^\circ\) and \(60^\circ\), respectively. If the height of the bridge from the banks is \( h \) metres and the width of the river is \( k \) metres, then \( h : k \) is equal to:

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For problems involving angles of depression or elevation, use the tangent function to relate the height and distance.
Updated On: Apr 19, 2025
  • \( 1 : 2\sqrt{3} \)
  • \( 4 : 3\sqrt{3} \)
  • \( \sqrt{3} : 2 \)
  • \( \sqrt{3} : 4 \)
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The Correct Option is A

Solution and Explanation

Let the height of the bridge be \( h \) metres and the width of the river be \( k \) metres. From the geometry of the situation, for the first angle of depression (\(30^\circ\)), we can use the tangent function: \[ \tan(30^\circ) = \frac{h}{k_1} \quad \Rightarrow \quad k_1 = \frac{h}{\tan(30^\circ)} = \frac{h}{\frac{1}{\sqrt{3}}} = h\sqrt{3} \] For the second angle of depression (\(60^\circ\)): \[ \tan(60^\circ) = \frac{h}{k_2} \quad \Rightarrow \quad k_2 = \frac{h}{\tan(60^\circ)} = \frac{h}{\sqrt{3}} = \frac{h}{\sqrt{3}} \] Thus, the ratio \( h : k \) is: \[ h : k = 1 : 2\sqrt{3} \] Therefore, the correct answer is \( 1 : 2\sqrt{3} \).
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