Q11

1 markMCQSection A

A car is moving away from the base of a 30 m high tower. The angle of elevation of the top of the tower from the car at an instant, when the car is $10\sqrt{3}$ m away from the base of the tower, is :

(a) 30° (b) 45° (c) 90° (d) 60°

Q: A car is moving away from the base of a 30 m high tower. The angle of elevation of the top of the tower from the car at an instant, when the car is $10\sqrt{3}$ m away from the base of the tower, is : (a) 30° (b) 45° (c) 90° (d) 60°

(d) 60° — $\tan\theta = \frac{30}{10\sqrt{3}} = \sqrt{3}$, and $\tan 60° = \sqrt{3}$, so $\theta = 60°$.

Some Applications of Trigonometry

Heights and Distances - Angle of Elevation

Options

(A)30°

(B)45°

(D)60°

Official Answer

(d) 60° — $\tan\theta = \frac{30}{10\sqrt{3}} = \sqrt{3}$ , and $\tan 60° = \sqrt{3}$ , so $\theta = 60°$ .

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Marking Scheme

11 mark: correct option (d) 60°; full credit reasoning: $\tan\theta = \frac{30}{10\sqrt{3}} = \sqrt{3} \implies \theta = 60°$ .

Hint

Form the right triangle with tower height as opposite and given distance as adjacent, then use $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ to find θ.

Quick Oral Answer

Using $\tan\theta = \frac{\text{height}}{\text{distance}} = \frac{30}{10\sqrt{3}} = \sqrt{3}$ , and since $\tan 60° = \sqrt{3}$ , the angle of elevation is 60°.

Analysis & Explanation

A standard height-and-distance application of trigonometric ratios in a right triangle.

Concept

The tower (height, opposite side), the ground distance (adjacent side), and the line of sight form a right triangle; $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$ .

Key points

$\tan\theta = \frac{30}{10\sqrt{3}} = \sqrt{3}$ , and since $\tan 60° = \sqrt{3}$ , $\theta = 60°$ — option (d).

Common mistakes

Not rationalising 30/(10√3) properly and misreading the resulting ratio.
Confusing angle of elevation with angle of depression.
Assuming $\theta = 90°$ by mistakenly treating the car as being at the base of the tower.

Real-world

This $\tan\theta = \frac{\text{height}}{\text{distance}}$ model is used to estimate building or tower heights from ground-level angle measurements in surveying.

Common Mistakes

1Not simplifying $\frac{30}{10\sqrt{3}}$ properly, leaving it as an unrecognized value instead of √3.
2Confusing angle of elevation (looking up) with angle of depression (looking down).
3Using the wrong trigonometric ratio, e.g., $\sin\theta$ or $\cos\theta$ instead of $\tan\theta$ , for a height/distance problem.

Interesting Facts

Heights and distances problems using trigonometry trace back to ancient surveying techniques; Hipparchus of Rhodes (2nd century BCE) is credited with early trigonometric tables used for such calculations.

The angle 60° here corresponds to one of the three standard angles (30°, 45°, 60°) that CBSE almost exclusively uses in heights-and-distances MCQs, since their tan values $\left(\frac{1}{\sqrt{3}}, 1, \sqrt{3}\right)$ are memorized values.

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Frequently Asked Questions

What is the formula for angle of elevation problems?

tan θ = height of the object (opposite side)/horizontal distance from the object (adjacent side), where θ is the angle of elevation.

Why is $\tan 60° = \sqrt{3}$ used here?

Because $\tan\theta = \frac{30}{10\sqrt{3}}$ simplifies to √3, and √3 is the standard tangent value for 60°, a value students memorize from the standard angle table.