Mathematics — CBSE Class 10 board question

Let the height of the kite from the ground = h m.

For Boy 1 (on ground):
String length = 60 m, angle of elevation = 30°

$$\sin 30° = \frac{h}{60} \implies \frac{1}{2} = \frac{h}{60} \implies h = 30 \text{ m}$$

∴ Height of kite from ground = 30 m

For Boy 2 (on roof of 20 m building):
Height of kite above Boy 2's level = 30 − 20 = 10 m
Angle of elevation = 45°

$$\tan 45° = \frac{10}{d} \implies 1 = \frac{10}{d} \implies d = 10 \text{ m}$$

where d = horizontal distance of kite from the building.

Total distance of Boy 1 from base of building:

Let horizontal distance of kite from Boy 1 = x.

$$\cos 30° = \frac{x}{60} \implies \frac{\sqrt{3}}{2} = \frac{x}{60} \implies x = 30\sqrt{3} \text{ m}$$

Since both boys are on opposite sides of the kite:

$$\text{Distance of Boy 1 from base of building} = x + d = 30\sqrt{3} + 10$$
$$= 30 \times 1.73 + 10 = 51.9 + 10 = \textbf{61.9 m}$$

Height of kite from ground = 30 m; Distance of Boy 1 from base of building = 61.9 m.

Source: Chapter 9, Section 9.1 Heights and Distances

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• Use sin for Boy 1 (string is hypotenuse, height is opposite side).
• For Boy 2, subtract building height from kite height to get the vertical difference, then use tan 45° = 1.
• Use cos 30° to find the horizontal distance from Boy 1 to the point below the kite.
• Since the boys are on opposite sides, add the two horizontal distances to get the total distance between Boy 1 and the building base.
• Examiners expect a clear diagram (optional but helpful), labelled equations, and substitution of √3 = 1.73 at the final step.