NCERT Class X Chapter 9: Some Application Of Trigonometry Example 7
NCERT Class X Chapter 9: Some Applications of Trigonometry Example 7
Question:
From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.
Given:
- Angles of depression to the two banks: \(30^\circ\) and \(45^\circ\)
- Height of the bridge above the banks: \(3\,\text{m}\)
To Find:
The width of the river.
Formula:
\[ \tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}} \]
Solution:
Step 1: Let the points where the lines of sight from the bridge meet the two banks be \(A\) and \(B\). Let \(P\) be the point on the bridge directly above the river at height \(3\,\text{m}\). Let the distances from \(P\) vertically down to \(A\) and \(B\) along the river banks be \(x\) and \(y\) respectively. The total width of the river is \(x + y\).
Step 2: For the bank with angle of depression \(45^\circ\), using the right triangle, we have:
\[ \tan 45^\circ = \frac{3}{x} \] \[ 1 = \frac{3}{x} \] \[ x = 3\,\text{m} \]Step 3: For the bank with angle of depression \(30^\circ\), using the right triangle, we have:
\[ \tan 30^\circ = \frac{3}{y} \] \[ \frac{1}{\sqrt{3}} = \frac{3}{y} \] \[ y = 3\sqrt{3}\,\text{m} \]Step 4: The width of the river is the sum of the two distances:
\[ \text{Width of the river} = x + y = 3 + 3\sqrt{3} \] \[ = 3(1 + \sqrt{3}) \] \[ \approx 3(1 + 1.732) \] \[ \approx 3 \times 2.732 \] \[ \approx 8.196\,\text{m} \]Result:
The width of the river is approximately \(8.2\,\text{m}\).
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