The angles of depression of the top and bottom of a 50m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building. (Use √(3) = 1.73)

MathsCBSE Class 10 · Basic · Sample paper, session 2025-26 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic sample paper 2025-26: question 34(B)

From the official CBSE sample question paper for 2025-26, 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 bottom of a 50m high building from the top of a tower are $4 5^{\circ}$ and $6 0^{\circ}$ respectively. Find the height of the tower and the horizontal distance between the tower and the building. (Use $3 = 1.73$ )

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

Draw a horizontal line from the top of the building to the tower; it cuts off a part of the tower of height 50 m.

Hint 2

Call the remaining part of the tower $h$ and the horizontal gap $x$ . Use $tan 4 5^{\circ}$ for the top of the building and $tan 6 0^{\circ}$ for its foot.

Hint 3

From $tan 4 5^{\circ}$ you get $h = x$ ; substitute this into $3 = \frac{h + 50}{x}$ and solve for $h$ .

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 $E C$ be the tower, with $E$ its top and $C$ its foot. Let $A B = 50$ m be the building, with $A$ its top and $B$ its foot. Draw $A D ⊥ E C$ , so $D C = A B = 50$ m. Let $E D = h$ m and $B C = A D = x$ m. From $E$ , the angle of depression of $A$ is $4 5^{\circ}$ , so $∠ E A D = 4 5^{\circ}$ . The angle of depression of $B$ is $6 0^{\circ}$ , so $∠ E B C = 6 0^{\circ}$ [alternate angles].
1 mark
In right $△ E D A$ : $tan 4 5^{\circ} = \frac{E D}{A D} = \frac{h}{x}$ , so $1 = \frac{h}{x}$ and hence $h = x$ .
1½ marks
In right $△ E C B$ : $tan 6 0^{\circ} = \frac{E C}{B C} = \frac{h + 50}{x}$ . Substituting $x = h$ gives $3 h = h + 50$ .
So $h (3 - 1) = 50$ , giving $h = \frac{50}{3 - 1} = \frac{50 ( 3 + 1 )}{2} = 25 (3 + 1)$ m.
1½ marks
Using $3 = 1.73$ : $h = 25 \times 2.73 = 68.25$ m. So the horizontal distance between the tower and the building is $x = h = 68.25$ m.
½ mark
Height of the tower $= E C = h + 50 = 68.25 + 50 = 118.25$ m.
½ mark

Answer: Height of tower = 118.25 m; horizontal distance = 68.25 m

Where the marks are

Correct labelled figure showing the tower, the building and both angles of depression (1 mark)
Uses $tan 4 5^{\circ}$ in $△ E D A$ to get $h = x$ (1½ marks)
Uses $tan 6 0^{\circ}$ in $△ E C B$ to get $h = 25 (3 + 1)$ m (1½ marks)
Horizontal distance = 68.25 m (½ mark)
Height of tower = 118.25 m (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 paperSample paper 2025-26 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic sample paper 2025-26: question 34(B)

Hints

Step-by-step solution

Revise the chapter

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