The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45° 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 2024, set 430/1/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2024: question 34(B)

Set in the CBSE Class 10 board exam of 2024, paper 430/1/1, section D, question 34(B), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(B) · 5 marks · Long answerSome Applications of Trigonometry

The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are $3 0^{\circ}$ and $4 5^{\circ}$ 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 2024 board paper, set 430/1/1.

Hints

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

Hint 1

Draw a horizontal line from the top of the 8 m building to the tall building; it cuts off a part of height (h − 8).

Hint 2

Angles of depression equal the angles of elevation at the lower points (alternate angles).

Hint 3

Use tan 45° in the big triangle and tan 30° in the upper small triangle, then solve the two equations.

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: AB = h m is the multi-storeyed building, CD = 8 m the small building, BD = x m the distance between them. Draw CE ⟂ AB, so EB = CD = 8 m, CE = BD = x and AE = (h − 8) m. Angle of depression of C is $3 0^{\circ}$ , so $∠ A C E = 3 0^{\circ}$ ; of D is $4 5^{\circ}$ , so $∠ A D B = 4 5^{\circ}$ [alternate angles].
1 mark
In right $△ A E C$ : $tan 3 0^{\circ} = \frac{A E}{C E} = \frac{h - 8}{x} \Rightarrow h - 8 = \frac{x}{3}$ ...(i)
1½ marks
In right $△ A B D$ : $tan 4 5^{\circ} = \frac{h}{x} \Rightarrow h = x$ ...(ii)
1 mark
From (i) and (ii): $x - \frac{x}{3} = 8 \Rightarrow x (3 - 1) = 83 \Rightarrow x = \frac{8 3 ( 3 + 1 )}{2} = 12 + 43 = 12 + 6.92 = 18.92$ m. So $h = x = 18.92$ m.
1½ marks

Answer: Height of multi-storeyed building = distance between buildings = $12 + 43$ = 18.92 m

Where the marks are

Correct labelled figure (1 mark)
$tan 3 0^{\circ} = \frac{h - 8}{x}$ giving $h - 8 = \frac{x}{3}$ (1½ marks)
$tan 4 5^{\circ}$ gives $h = x$ (1 mark)
Solves to $h = x = 12 + 43 = 18.92$ m (1½ marks)

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 paper2024 board paper, set 430/1/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2024: question 34(B)

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry