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

Given: Balloon of radius $r$ subtends angle 60° at observer's eye O. Angle of elevation of centre C is 45°.

To prove: Height of centre = $\sqrt{2}\,r$

Construction/Setup:

Let O be the eye of the observer and C be the centre of the balloon. Draw tangents OA and OB to the balloon from O.

Since the balloon subtends an angle of 60° at O:
$$\angle AOB = 60° \implies \angle AOC = 30°$$

In right $\triangle OAC$ (OA is tangent, so $\angle OAC = 90°$):
$$\sin(\angle AOC) = \frac{AC}{OC}$$
$$\sin 30° = \frac{r}{OC}$$
$$\frac{1}{2} = \frac{r}{OC} \implies OC = 2r$$

Finding height of centre C:

Let M be the point on the ground directly below O (observer's level), and let CM be the height of C above observer's level (taken as horizontal reference). The angle of elevation of C is 45°, so:

$$\sin 45° = \frac{h}{OC}$$

where $h$ = height of centre C above the observer's eye.

$$\frac{1}{\sqrt{2}} = \frac{h}{2r}$$
$$h = \frac{2r}{\sqrt{2}} = \sqrt{2}\,r$$

Hence, the height of the centre of the balloon is $\sqrt{2}$ times its radius. $\blacksquare$

Source: Chapter 9 — Some Applications of Trigonometry, Heights and Distances

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• The key step is recognising that $OA \perp AC$ (radius to tangent point is perpendicular), letting you use $\sin 30° = r/OC$ to get $OC = 2r$.
• Then use $\sin 45° = h/OC$ (not $\tan$, because $OC$ is a slant distance from observer to centre, not a horizontal base).
• Examiners expect a clear diagram description, correct triangle identification, and all trigonometric steps shown explicitly.
• Avoid confusing $\tan$ (used when base is known) with $\sin$ (used when hypotenuse is known); here $OC$ is the hypotenuse of the elevation triangle.