NCERT Class X Chapter 9: Some Application Of Trigonometry Example 2
NCERT Class X Chapter 9: Some Applications of Trigonometry Example 2
Question:
An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (You may take \( \sqrt{3} = 1.73 \))
Given:
Height of the pole = 5 m
Point to be reached is 1.3 m below the top, so height to reach = \( 5 - 1.3 = 3.7 \) m
Angle of inclination of ladder with horizontal = \( 60^\circ \)
\( \sqrt{3} = 1.73 \)
To Find:
1. The length of the ladder required.
2. The distance from the foot of the pole to the foot of the ladder.
Formula:
Trigonometric ratios in a right-angled triangle:
\[ \sin \theta = \frac{\text{Opposite side}}{\text{Hypotenuse}} \] \[ \cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}} \]
Solution:
Step 1: Let the length of the ladder be \( l \) and the distance from the foot of the pole to the foot of the ladder be \( x \).
Step 2: In the right-angled triangle, the vertical side (height to be reached) is 3.7 m, the hypotenuse is the ladder, and the angle with the horizontal is \( 60^\circ \).
Step 3: Using the sine ratio:
\[ \sin 60^\circ = \frac{3.7}{l} \] \[ \sin 60^\circ = \frac{\sqrt{3}}{2} \] \[ \frac{\sqrt{3}}{2} = \frac{3.7}{l} \] \[ l = \frac{3.7 \times 2}{\sqrt{3}} \] \[ l = \frac{7.4}{1.73} \] \[ l \approx 4.28 \text{ m} \]Step 4: Using the cosine ratio to find the distance from the foot of the pole to the foot of the ladder (\( x \)):
\[ \cos 60^\circ = \frac{x}{l} \] \[ \cos 60^\circ = \frac{1}{2} \] \[ \frac{1}{2} = \frac{x}{4.28} \] \[ x = 4.28 \times \frac{1}{2} \] \[ x = 2.14 \text{ m} \]Result:
The length of the ladder should be approximately 4.28 m.
The foot of the ladder should be placed approximately 2.14 m away from the foot of the pole.
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