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Some Applications Of Trigonometry
EXERCISE 9.1
CBSE_NCERT_TEXTBOOK
Grade 10
Question:
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30° (see Fig. 9.11).
Step-by-Step Solution
Key Concept: In a right‑angled triangle, the sine of an acute angle equals the ratio of the length of the side opposite the angle to the length of the hypotenuse. Hence, height = (hypotenuse) × sin θ.
1. Draw the right‑angled triangle formed by the pole, the ground and the rope.\
• Let \(AB\) be the pole (vertical side), \(BC\) the ground (horizontal side) and \(AC\) the rope (hypotenuse).\
• Given: \(AC = 20\,\text{m}\) and \(\angle B = 30^{\circ}\) (angle between rope and ground).\
2. Identify the required side: the height of the pole is \(AB\), which is opposite \(\angle B\).\
3. Use the definition of sine for \(\angle B\):\
$$\sin 30^{\circ} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC}$$\
4. Substitute the known values:\
$$\sin 30^{\circ} = \frac{AB}{20}\
\Rightarrow AB = 20 \times \sin 30^{\circ}$$\
5. Evaluate \(\sin 30^{\circ} = \frac{1}{2}\):\
$$AB = 20 \times \frac{1}{2} = 10\,\text{m}$$\
6. Hence, the height of the pole is \(10\) metres.
• Let \(AB\) be the pole (vertical side), \(BC\) the ground (horizontal side) and \(AC\) the rope (hypotenuse).\
• Given: \(AC = 20\,\text{m}\) and \(\angle B = 30^{\circ}\) (angle between rope and ground).\
2. Identify the required side: the height of the pole is \(AB\), which is opposite \(\angle B\).\
3. Use the definition of sine for \(\angle B\):\
$$\sin 30^{\circ} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AB}{AC}$$\
4. Substitute the known values:\
$$\sin 30^{\circ} = \frac{AB}{20}\
\Rightarrow AB = 20 \times \sin 30^{\circ}$$\
5. Evaluate \(\sin 30^{\circ} = \frac{1}{2}\):\
$$AB = 20 \times \frac{1}{2} = 10\,\text{m}$$\
6. Hence, the height of the pole is \(10\) metres.
Correct Answer: 10 m
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