NCERT Class X Chapter 9: Some Application Of Trigonometry Exercise 9.1 Question 11
NCERT Class X Chapter 9: Some Applications of Trigonometry
Question:
A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.
Given:
Let the height of the tower be \( h \) meters and the width of the canal be \( x \) meters.
The angle of elevation from the point on the other bank directly opposite the tower is \( 60^\circ \).
From another point 20 m farther on the same line, the angle of elevation is \( 30^\circ \).
To Find:
Height of the tower (\( h \)) and width of the canal (\( x \)).
Formula:
For a right triangle,
\[
\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}
\]
Solution:
Step 1: Let the height of the tower be \( h \) m and the width of the canal be \( x \) m. From the first point (directly opposite the tower), the angle of elevation is \( 60^\circ \).
\[ \tan 60^\circ = \frac{h}{x} \] \[ \sqrt{3} = \frac{h}{x} \] \[ h = x\sqrt{3} \tag{1} \]Step 2: From the second point, which is 20 m farther from the tower, the angle of elevation is \( 30^\circ \). The distance from this point to the foot of the tower is \( x + 20 \) m.
\[ \tan 30^\circ = \frac{h}{x + 20} \] \[ \frac{1}{\sqrt{3}} = \frac{h}{x + 20} \] \[ h = \frac{x + 20}{\sqrt{3}} \tag{2} \]Step 3: Equate the two expressions for \( h \) from (1) and (2):
\[ x\sqrt{3} = \frac{x + 20}{\sqrt{3}} \] \[ x\sqrt{3} \times \sqrt{3} = x + 20 \] \[ 3x = x + 20 \] \[ 3x - x = 20 \] \[ 2x = 20 \] \[ x = 10 \]Step 4: Substitute \( x = 10 \) into equation (1) to find \( h \):
\[ h = x\sqrt{3} = 10\sqrt{3} \]Result:
Height of the tower \( = 10\sqrt{3} \) m
Width of the canal \( = 10 \) m
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