From the top of a 9 m high building, the angle of elevation of the top of a cable tower is 60° and angle of depression of its foot is 45°. Determine the height of the tower and distance between building and tower. (Use √(3) = 1.732)

MathsCBSE Class 10 · Basic · Sample paper, session 2024-25 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic sample paper 2024-25: question 35(A)

From the official CBSE sample question paper for 2024-25, section D, question 35(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35(A) · 5 marks · Long answerSome Applications of Trigonometry

From the top of a 9 m high building, the angle of elevation of the top of a cable tower is $6 0^{\circ}$ and angle of depression of its foot is $4 5^{\circ}$ . Determine the height of the tower and distance between building and tower. (Use $3 = 1.732$ )

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

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 splits the tower into a 9 m part and an unknown part $h$ .

Hint 2

Use $tan 4 5^{\circ}$ to find the horizontal distance first, then use $tan 6 0^{\circ}$ to find $h$ .

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: AB is the building with $A B = 9$ m, and B is its top. CD is the tower with foot C and top D. Draw $B E ⊥ C D$ , so $E C = A B = 9$ m and $B E = C A = x$ . Let $D E = h$ .
Angle of elevation of D from B is $∠ D B E = 6 0^{\circ}$ . Angle of depression of C from B is $∠ E B C = 4 5^{\circ} = ∠ B C A$ .
1 mark
In right $△ D E B$ : $tan 6 0^{\circ} = \frac{D E}{B E} \Rightarrow 3 = \frac{h}{x}$
1 mark
$\Rightarrow h = x 3$ …(i)
½ mark
In right $△ B A C$ : $tan 4 5^{\circ} = \frac{A B}{A C} \Rightarrow 1 = \frac{9}{x}$
1 mark
$\Rightarrow x = 9$ m …(ii). This is the distance between the building and the tower.
½ mark
From (i) and (ii): $h = 9 \times 1.732 = 15.588$ m
½ mark
Height of tower $C D = h + 9 = 15.588 + 9 = 24.588$ m
½ mark

Answer: Height of tower = 24.588 m; distance between building and tower = 9 m

Where the marks are

Correct labelled figure (1 mark)
Uses $tan 6 0^{\circ} = \frac{h}{x}$ (1 mark)
Gets $h = x 3$ (½ mark)
Uses $tan 4 5^{\circ} = \frac{9}{x}$ (1 mark)
Gets $x = 9$ m (½ mark)
$h = 15.588$ m (½ mark)
Tower height = 24.588 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 2024-25 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic sample paper 2024-25: question 35(A)

Hints

Step-by-step solution

Revise the chapter

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