Mathematics — Standard (041) — CBSE Class 10 board question

Diagram: Let the ground be at point A. The kite is at point K, 60 m above the ground. The building stands vertically; roof is at point R, ground level at point B (foot of building).

Let height of roof = h m. Height of kite above roof = (60 − h) m.
Horizontal distance between kite and building = d m.

From the bottom of the building (point B), angle of elevation = 45°:

$$\tan 45° = \frac{60}{d}$$
$$1 = \frac{60}{d} \implies d = 60 \text{ m}$$

From the roof (point R), angle of elevation = 30°:

$$\tan 30° = \frac{60 - h}{d}$$
$$\frac{1}{\sqrt{3}} = \frac{60 - h}{60}$$
$$60 - h = \frac{60}{\sqrt{3}} = 20\sqrt{3}$$
$$h = 60 - 20\sqrt{3} = 60 - 20 \times 1.73 = 60 - 34.6 = 25.4 \text{ m}$$

Length of string (from roof R to kite K):

$$\sin 30° = \frac{60 - h}{\text{string}}$$
$$\frac{1}{2} = \frac{20\sqrt{3}}{\text{string}}$$
$$\text{String} = 40\sqrt{3} = 40 \times 1.73 = 69.2 \text{ m}$$

∴ Height of roof = 25.4 m; Length of string = 69.2 m.

Source: Chapter 9 — Heights and Distances, Section 9.1

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• Two right triangles are formed: one from the ground (angle 45°) and one from the roof (angle 30°), both looking at the same kite at 60 m height.
• From the ground triangle, find horizontal distance first (tan 45° = 1 makes it straightforward).
• Use that distance in the roof triangle to find the roof height.
• For the string length, use sin 30° with the vertical height difference (60 − h) as the opposite side — the string is the hypotenuse from Ravi's position to the kite.
• Always substitute √3 = 1.73 numerically at the end as the question requires.