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Some Applications of Trigonometry
CH09 Question Bank
CBSE_CH09_QUESTION_BANK
Grade 10

Question:

[Case Study]

A kite enthusiast is flying a kite at a park. The kite string is stretched taut and makes an angle of $60^\circ$ with the horizontal ground. The length of the string released is $100$ m. Simultaneously, a drone hovering at a fixed height observes the kite.

(a) Find the vertical height of the kite above the ground. [1 Mark]
(b) Find the horizontal distance of the kite from the person holding the string. [1 Mark]
(c) Verify the Pythagorean relationship $h^2 + d^2 = L^2$ for this kite position. [1 Mark]
(d) If the string angle decreases to $30^\circ$ while maintaining the string length at $100$ m, what will be the new height of the kite? [1 Mark]

Step-by-Step Solution

Key Concept: Case study on applications of trigonometry (heights and distances).
(a) Find the vertical height of the kite above the ground. [1 Mark]
$\sin60^\circ = \dfrac{h}{100} \Rightarrow \dfrac{\sqrt3}{2} = \dfrac{h}{100} \Rightarrow h = 50\sqrt3$ m. [1.0 Mark]

(b) Find the horizontal distance of the kite from the person holding the string. [1 Mark]
$\cos60^\circ = \dfrac{d}{100} \Rightarrow \dfrac{1}{2} = \dfrac{d}{100} \Rightarrow d = 50$ m. [1.0 Mark]

(c) Verify the Pythagorean relationship $h^2 + d^2 = L^2$ for this kite position. [1 Mark]
$h^2 + d^2 = (50\sqrt3)^2 + 50^2 = 7500 + 2500 = 10000 = 100^2 = L^2$. Verified. [1.0 Mark]

(d) If the string angle decreases to $30^\circ$ while maintaining the string length at $100$ m, what will be the new height of the kite? [1 Mark]
$h_{\text{new}} = 100 \sin30^\circ = 100 \times \dfrac{1}{2} = 50$ m. [1.0 Mark]

Correct Answer: $\sin60^\circ = \dfrac{h}{100} \Rightarrow \dfrac{\sqrt3}{2} = \dfrac{h}{100} \Rightarrow h = 50\sqrt3$ m. [1.0 Mark] | $\cos60^\circ = \dfrac{d}{100} \Rightarrow \dfrac{1}{2} = \dfrac{d}{100} \Rightarrow d = 50$ m. [1.0 Mark] | $h^2 + d^2 = (50\sqrt3)^2 + 50^2 = 7500 + 2500 = 10000 = 100^2 = L^2$. Verified. [1.0 Mark] | $h_{\text{new}} = 100 \sin30^\circ = 100 \times \dfrac{1}{2} = 50$ m. [1.0 Mark]
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