From the top of a building 60 m high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60° respectively. Find the height of the tower. Also, find the distance between the building and the tower. (Use √(3) = 1·732)

MathsCBSE Class 10 · Basic · Board exam 2023, set 430/2/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2023: question 32(a)

Set in the CBSE Class 10 board exam of 2023, paper 430/2/1, section D, question 32(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(a) · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry

From the top of a building 60 m high, the angles of depression of the top and bottom of a tower are observed to be $3 0^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the tower. Also, find the distance between the building and the tower. (Use $3 = 1 \cdot 732$ )

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 430/2/1.

Hints

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

Hint 1

Draw a horizontal line from the top of the tower to the building; it cuts off a piece of length (60 − h) at the top.

Hint 2

Use the 60° angle with the full building height to find the distance x first.

Hint 3

Then use the 30° angle in the small top triangle with side (60 − h).

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 AB = 60 m be the building (A at the foot, B at the top) and CD = h m be the tower (C at the foot, D at the top), with AC = x m. Draw DE ⟂ AB, so DE = AC = x and AE = h, BE = 60 − h. Angle of depression of C from B is 60°, so ∠BCA = 60°; angle of depression of D from B is 30°, so ∠BDE = 30° (alternate angles).
1 mark
In right $△ B A C$ : $tan 6 0^{\circ} = \frac{A B}{A C} = \frac{60}{x} \Rightarrow 3 = \frac{60}{x} \Rightarrow x = \frac{60}{3} = 203$ m ...(i)
1½ marks
In right $△ B E D$ : $tan 3 0^{\circ} = \frac{B E}{D E} = \frac{60 - h}{x} \Rightarrow 60 - h = \frac{x}{3} = \frac{20 3}{3} = 20$ ...(ii)
1½ marks
Distance between building and tower $= x = 203 = 20 \times 1.732 = 34.64$ m
½ mark
From (ii), $h = 60 - 20 = 40$ m. Height of the tower = 40 m.
½ mark

Answer: Height of tower = 40 m; distance between building and tower = $203$ = 34.64 m

Where the marks are

Correct labelled figure (1 mark)
Uses $tan 6 0^{\circ} = \frac{60}{x}$ to get $x = 203$ (1½ marks)
Uses $tan 3 0^{\circ} = \frac{60 - h}{x}$ to get $60 - h = 20$ (1½ marks)
Distance = 34.64 m (1/2 mark)
Height of tower = 40 m (1/2 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 paper2023 board paper, set 430/2/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2023: question 32(a)

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry