MathsCBSE Class 10 · Basic · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic sample paper 2026-27: question 35(B)

From the official CBSE sample question paper for 2026-27, section D, question 35(B), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(B) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry

A statue standing on a pedestal, is 1.6 m tall. From a point on the ground, the angles of elevation of the top and bottom of the statue are $6 0^{\circ}$ and $4 5^{\circ}$ respectively. Find the height of the pedestal.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

Draw the vertical line through the pedestal and statue. Note that the bottom of the statue is at height $h$ and the top is at height $h + 1.6$ .

Hint 2

Both right triangles share the same horizontal distance AB. Write AB from each triangle using tan.

Hint 3

Use $tan 4 5^{\circ} = 1$ and $tan 6 0^{\circ} = 3$ , set the two expressions for AB equal, and rationalise the denominator.

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.

Figure: Let A be the point on the ground, BC the pedestal of height $h$ m, and CD the statue of height 1.6 m, with B the foot of the pedestal and $∠ A B C = 9 0^{\circ}$ . The angle of elevation of C (bottom of the statue) from A is $∠ B A C = 4 5^{\circ}$ . The angle of elevation of D (top of the statue) from A is $∠ B A D = 6 0^{\circ}$ . Then $B D = h + 1.6$ m.
1 mark
In right $△ A B C$ : $tan 4 5^{\circ} = \frac{B C}{A B} \Rightarrow 1 = \frac{h}{A B} \Rightarrow A B = h$ ...(i)
1 mark
In right $△ A B D$ : $tan 6 0^{\circ} = \frac{B D}{A B} \Rightarrow 3 = \frac{h + 1.6}{A B} \Rightarrow A B = \frac{h + 1.6}{3}$ ...(ii)
1 mark
Equating (i) and (ii): $h = \frac{h + 1.6}{3} \Rightarrow 3 h = h + 1.6$
½ mark
$(3 - 1) h = 1.6 \Rightarrow h = \frac{1.6}{3 - 1} \times \frac{3 + 1}{3 + 1} = \frac{1.6 ( 3 + 1 )}{2} = 0.8 (3 + 1)$
1 mark
Hence the height of the pedestal is $0.8 (3 + 1)$ m $\approx 2.19$ m.
½ mark

Answer: Height of pedestal = $0.8 (3 + 1)$ m ≈ 2.19 m

Where the marks are

Correct labelled figure with both angles and heights (1 mark)
Uses $tan 4 5^{\circ}$ to get $A B = h$ (1 mark)
Uses $tan 6 0^{\circ}$ to get $A B = \frac{h + 1.6}{3}$ (1 mark)
Equates the two expressions for AB (½ mark)
Solves to get $h = \frac{1.6}{3 - 1} = 0.8 (3 + 1)$ (1 mark)
States the height of the pedestal with unit (½ 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 2026-27 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic sample paper 2026-27: question 35(B)

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry