India gate (formerly known as All India war memorial) is located near Karthavya path (formerly Rajpath) at New Delhi. It stands as a memorial to 74187 soldiers of Indian Army, who gave their life in the first world war. This 42m tall structure was designed by Sir Edwin Lutyens in the style of Roman triumphal arches. A student Shreya of height 1 m visited India Gate as a part of her study tour. i. What is the angle of elevation from Shreya’s eye to the top of India Gate, if she is standing at a distance of 41m away from the India Gate? ii. If Shreya observes the angle of elevation from her eye to the top of India Gate to be 60°, then how far is the she standing from the base of the India Gate? iii. (A) If the angle of elevation from Shreya’s eye changes from 45° to 30°, when she moves some distance back from the original position. Find the distance she moves back. iii. (B) If Shreya moves to a point which is at a distance of (41)/(√(3)) m from the India Gate, then find the angle of elevation made by her eye to the top of India Gate.

(i) 45°; (ii) (413)/(3) m (≈ 23.67 m); (iii)(a) 41(3 - 1) m (≈ 30.01 m), (iii)(b) 60°

MathsCBSE Class 10 · Standard · Sample paper, session 2025-26 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard sample paper 2025-26: question 38

From the official CBSE sample question paper for 2025-26, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry

India gate (formerly known as All India war memorial) is located near Karthavya path (formerly Rajpath) at New Delhi. It stands as a memorial to 74187 soldiers of Indian Army, who gave their life in the first world war. This 42m tall structure was designed by Sir Edwin Lutyens in the style of Roman triumphal arches. A student Shreya of height 1 m visited India Gate as a part of her study tour.

i.
What is the angle of elevation from Shreya’s eye to the top of India Gate, if she is standing at a distance of 41m away from the India Gate?
1 mark
ii.
If Shreya observes the angle of elevation from her eye to the top of India Gate to be 60°, then how far is the she standing from the base of the India Gate?
1 mark
iii.
(A) If the angle of elevation from Shreya’s eye changes from 45° to 30°, when she moves some distance back from the original position. Find the distance she moves back.
2 marks
OR
iii.
(B) If Shreya moves to a point which is at a distance of $\frac{41}{3}$ m from the India Gate, then find the angle of elevation made by her eye to the top of India Gate.
2 marks

Figure in the paper: Photograph of India Gate, a tall arched memorial structure on a road. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

Open one hint at a time and try again before the next.

Hint 1

Measure the height from Shreya's eye, not from the ground: subtract her 1 m height.

Hint 2

Use $tan$ = opposite / adjacent in the right triangle at her eye level.

Hint 3

For (iii)(a), the new distance is 41 m plus the distance she moves back.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Shreya's eye is 1 m above the ground, so the height of the top above her eye is $42 - 1 = 41$ m. Her horizontal distance from the gate is 41 m.
$tan θ = \frac{41}{41} = 1$
½ mark
(i) $tan θ = 1$ , so $θ = 4 5^{\circ}$ .
½ mark
(ii) Let the distance be $x$ m. The height above her eye is still 41 m.
$tan 6 0^{\circ} = \frac{41}{x}$
½ mark
(ii) $3 = \frac{41}{x}$
$\Rightarrow x = \frac{41}{3} = \frac{41 3}{3}$ m $\approx 23.67$ m
½ mark
(iii)(a) At 45° she stands 41 m away (from part (i)). Let her move back $x$ m, so her new distance is $(41 + x)$ m.
$tan 3 0^{\circ} = \frac{41}{41 + x}$
$\Rightarrow \frac{1}{3} = \frac{41}{41 + x}$
1 mark
(iii)(a) $41 + x = 413$
$\Rightarrow x = 41 (3 - 1)$ m $\approx 30.01$ m
1 mark
(iii)(b) Height above her eye is 41 m and distance is $\frac{41}{3}$ m.
$tan θ = \frac{41}{41/ 3} = 3$
1 mark
(iii)(b) $tan θ = 3$ , so $θ = 6 0^{\circ}$ .
1 mark

Answer: (i) 45°; (ii) $\frac{41 3}{3}$ m (≈ 23.67 m); (iii)(a) $41 (3 - 1)$ m (≈ 30.01 m), (iii)(b) 60°

Where the marks are

(i) Uses height 41 m above eye level with $tan θ = 1$ (½ mark); $θ = 4 5^{\circ}$ (½ mark)
(ii) Writes $tan 6 0^{\circ} = 41/ x$ (½ mark); $x = \frac{41 3}{3}$ m (½ mark)
(iii)(a) Writes $tan 3 0^{\circ} = \frac{41}{41 + x}$ (1 mark); $x = 41 (3 - 1)$ m (1 mark)
(iii)(b) Gets $tan θ = 3$ (1 mark); $θ = 6 0^{\circ}$ (1 mark)

Revise the chapter

Chapter: Some Applications of Trigonometry (notes, formulas, where marks are lost and every past question)
Topic: Line of Sight and Angle of Elevation

Practise this chapterBoard-style questions from Some Applications of Trigonometry, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2025-26 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2025-26: question 38

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry