The angles of depression of the top and the foot of a 9 m tall building from the top of a multi-storeyed building are 30° and 60° respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use √(3) = 1.73)

MathsCBSE Class 10 · Basic · Board exam 2025, set 430/6/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2025: question 34

Set in the CBSE Class 10 board exam of 2025, paper 430/6/1, section D, question 34, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry

The angles of depression of the top and the foot of a 9 m tall building from the top of a multi-storeyed building are $3 0^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the multi-storeyed building and the distance between the two buildings. (Use $3 = 1.73$ )

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/6/1.

Hints

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

Hint 1

Draw a horizontal line from the top of the small building to the tall building; the top part has height h − 9.

Hint 2

Write tan 30° in the upper small triangle and tan 60° in the large triangle, then solve the two equations together.

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 CD be the multi-storeyed building of height $h$ m (D at top, C at foot) and AB the 9 m building (B top, A foot), with $A C = x$ m. Draw BE ⟂ CD, so $E C = 9$ m, $D E = (h - 9)$ m, $B E = A C = x$ . Angles of depression from D: to B is $3 0^{\circ}$ , to A is $6 0^{\circ}$ ; so $∠ D B E = 3 0^{\circ}$ and $∠ D A C = 6 0^{\circ}$ [alternate angles].
1 mark
In right $△ D E B$ : $tan 3 0^{\circ} = \frac{D E}{B E} = \frac{h - 9}{x} \Rightarrow \frac{1}{3} = \frac{h - 9}{x} \Rightarrow x = (h - 9) 3$ …(i)
1½ marks
In right $△ D C A$ : $tan 6 0^{\circ} = \frac{D C}{A C} = \frac{h}{x} \Rightarrow h = x 3$ …(ii)
1½ marks
Substituting (ii) in (i): $x = (x 3 - 9) 3 = 3 x - 93 \Rightarrow 2 x = 93 \Rightarrow x = 4.5 3 = 4.5 \times 1.73 \approx 7.79$ m.
Then $h = x 3 = 4.5 \times 3 = 13.5$ m.
Height of the multi-storeyed building = 13.5 m; distance between the buildings ≈ 7.79 m.
1 mark

Answer: Height = 13.5 m; distance between buildings ≈ 7.79 m

Where the marks are

Correct labelled figure (1 mark)
Uses $tan 3 0^{\circ} = \frac{h - 9}{x}$ to get $x = (h - 9) 3$ (1½ marks)
Uses $tan 6 0^{\circ} = \frac{h}{x}$ to get $h = x 3$ (1½ marks)
Solves: $x \approx 7.79$ m and $h = 13.5$ m (½ + ½ mark)

Revise the chapter

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

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 paper2025 board paper, set 430/6/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2025: question 34

Hints

Step-by-step solution

Revise the chapter

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