Kite festival is celebrated in many countries at different times of the year. In India, every year 14th January is celebrated as International Kite Day. On this day many people visit India and participate in the festival by flying various kinds of kites. The picture given below, shows three kites flying together. In Fig. 5, the angles of elevation of two kites (Points A and B) from the hands of a man (Point C) are found to be 30° and 60° respectively. Taking AD = 50 m and BE = 60 m.
Case Study - 1 : Kite Festival. In Fig. 5, the angles of elevation of two kites (Points A and B) from the hands of a man (Point C) are found to be 30° and 60° respectively. Taking AD = 50 m and BE = 60 m, find:
Generated by claude-sonnet-4-6 · 2026-06-15 10:37 · grounding stimulus
Model Answer
(1) Length of strings CA and CB:
For kite A: angle of elevation = 30°, AD = 50 m
$$\sin 30° = \frac{AD}{CA} \Rightarrow \frac{1}{2} = \frac{50}{CA} \Rightarrow CA = 100 \text{ m}$$
For kite B: angle of elevation = 60°, BE = 60 m
$$\sin 60° = \frac{BE}{CB} \Rightarrow \frac{\sqrt{3}}{2} = \frac{60}{CB} \Rightarrow CB = \frac{120}{\sqrt{3}} = 40\sqrt{3} \text{ m}$$
(2) Distance 'd' between the two kites:
$$CD = \frac{AD}{\tan 30°} = \frac{50}{1/\sqrt{3}} = 50\sqrt{3} \text{ m}$$
$$CE = \frac{BE}{\tan 60°} = \frac{60}{\sqrt{3}} = 20\sqrt{3} \text{ m}$$
Since D, C, E are collinear and A, B are above D and E respectively:
$$AB = d = \sqrt{(AD - BE)^2 + (DC + CE)^2}$$
$$= \sqrt{(50-60)^2 + (50\sqrt{3}+20\sqrt{3})^2} = \sqrt{100 + (70\sqrt{3})^2}$$
$$= \sqrt{100 + 14700} = \sqrt{14800} = 20\sqrt{37} \text{ m}$$
Source: Chapter 9 – Some Applications of Trigonometry
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Explanation
- For string length, use sin θ = perpendicular/hypotenuse (hypotenuse = string = line of sight).
- For horizontal distance (DC, CE), use tan θ = perpendicular/base.
- Since D and E are on opposite sides of C, total horizontal separation = DC + CE.
- Use Pythagoras for the slant distance AB, as A and B are at different heights.
- Key values: sin 30° = ½, sin 60° = √3/2, tan 30° = 1/√3, tan 60° = √3.