Question 111 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 10310\sqrt{3} m away from the base of the tower, is :

(a) 30° (b) 45° (c) 90° (d) 60°

Some Applications of Trigonometry
Heights and Distances - Angle of Elevation

Options

(A)30°
(B)45°
(C)90°
(D)60°

Official answer

(d) 60° — tan⁡θ=30103=3\tan\theta = \frac{30}{10\sqrt{3}} = \sqrt{3}, and tan⁡60°=3\tan 60° = \sqrt{3}, so θ=60°\theta = 60°.

Answer keywords:angle of elevationheight and distancetan thetatower height 30 m60 degrees

Marking scheme

  1. 1.1 mark: correct option (d) 60°; full credit reasoning: tan⁡θ=30103=3  ⟹  θ=60°\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⁡θ=oppositeadjacent\tan\theta = \frac{\text{opposite}}{\text{adjacent}} to find θ.

Quick oral answer

Using tan⁡θ=heightdistance=30103=3\tan\theta = \frac{\text{height}}{\text{distance}} = \frac{30}{10\sqrt{3}} = \sqrt{3}, and since tan⁡60°=3\tan 60° = \sqrt{3}, the angle of elevation is 60°.

Common mistakes

  1. 1.Not simplifying 30103\frac{30}{10\sqrt{3}} properly, leaving it as an unrecognized value instead of √3.
  2. 2.Confusing angle of elevation (looking up) with angle of depression (looking down).
  3. 3.Using the wrong trigonometric ratio, e.g., sin⁡θ\sin\theta or cos⁡θ\cos\theta instead of tan⁡θ\tan\theta, for a height/distance problem.

Analysis and 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⁡θ=oppositeadjacent\tan\theta = \frac{\text{opposite}}{\text{adjacent}}.

Key points

  • tan⁡θ=30103=3\tan\theta = \frac{30}{10\sqrt{3}} = \sqrt{3}, and since tan⁡60°=3\tan 60° = \sqrt{3}, θ=60°\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 θ=90°\theta = 90° by mistakenly treating the car as being at the base of the tower.

Real-world

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

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 (13,1,3)\left(\frac{1}{\sqrt{3}}, 1, \sqrt{3}\right) are memorized values.

Question details

Q11 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Some Applications of Trigonometry
Topic
Heights and Distances - Angle of Elevation
Bloom's level
apply
Syllabus topic
Unit 4: Trigonometry — Some Applications of Trigonometry: Heights and Distances
Author
CBSE
Reviewed by
Padhantu Academic Team

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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°=3\tan 60° = \sqrt{3} used here?

Because tan⁡θ=30103\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.