A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from the point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower.

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

CBSE Class 10 Mathematics Basic 2023: question 35(a)

Set in the CBSE Class 10 board exam of 2023, paper 430/1/1, 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 answerCompetency-basedSome Applications of Trigonometry

A TV tower stands vertically on the bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $6 0^{\circ}$ . From another point 20 m away from the point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is $3 0^{\circ}$ . Find the height of the tower.

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

Hints

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

Hint 1

Draw the tower and mark both observation points on the same straight line through its foot. The farther point is 20 m beyond the nearer one.

Hint 2

Let the canal width be $x$ . Write the tangent of each angle as height over horizontal distance.

Hint 3

From $tan 6 0^{\circ}$ you get $h = 3 x$ . Put this into the $tan 3 0^{\circ}$ equation and solve for $x$ .

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 = $h$ m be the tower, with foot A and top B. C is the point on the opposite bank, with AC = $x$ m and $∠ B C A = 6 0^{\circ}$ . D is 20 m beyond C on line AC, so AD = $(20 + x)$ m and $∠ B D A = 3 0^{\circ}$ . Also $∠ B A D = 9 0^{\circ}$ .
1 mark
In right $△ B A C$ : $tan 6 0^{\circ} = \frac{A B}{A C} = \frac{h}{x}$ , so $3 = \frac{h}{x}$ , i.e. $h = 3 x$ ...(1)
1½ marks
In right $△ B A D$ : $tan 3 0^{\circ} = \frac{A B}{A D} = \frac{h}{20 + x}$ , so $\frac{1}{3} = \frac{h}{20 + x}$ , i.e. $3 h = 20 + x$ ...(2)
1½ marks
Substitute (1) in (2): $3 (3 x) = 20 + x$ , so $3 x - x = 20$ , giving $x = 10$ m.
½ mark
From (1), $h = 3 \times 10 = 103$ m $\approx 17.32$ m. Hence the height of the tower is $103$ m.
½ mark

Answer: Height of tower = $103$ m ≈ 17.32 m (the width of the canal is 10 m)

Where the marks are

Correct labelled figure (1 mark)
Uses $tan 6 0^{\circ} = h / x$ to get $h = 3 x$ (1½ marks)
Uses $tan 3 0^{\circ} = h / (20 + x)$ to get $3 h = 20 + x$ (1½ marks)
Solves to get $x = 10$ (½ mark)
Height $= 103$ m (½ mark)

Revise the chapter

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

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 430/1/1 (Basic): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Basic 2023: question 35(a)

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry