A pole 6 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point P on the ground is 60° and the angle of depression of the point P from the top of the tower is 45°. Find the height of the tower and the distance of point P from the foot of the tower. (Use √(3)=1.73)

MathsCBSE Class 10 · Standard · Board exam 2024, set 30/1/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2024: question 35

Set in the CBSE Class 10 board exam of 2024, paper 30/1/1, section D, question 35, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 5 marks · Long answerCompetency-basedSome Applications of Trigonometry

A pole 6 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point P on the ground is $6 0^{\circ}$ and the angle of depression of the point P from the top of the tower is $4 5^{\circ}$ . Find the height of the tower and the distance of point P from the foot of the tower. (Use $3 = 1.73$ )

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/1/1.

Hints

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

Hint 1

The angle of depression from the top of the tower to P equals the angle of elevation of the tower top from P.

Hint 2

Let the tower be h and the distance be x. Write tan 45° using the tower only, and tan 60° using tower plus pole.

Hint 3

From tan 45° you get h = x. Substitute this into h + 6 = √3 x and rationalise.

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: Draw a vertical line $A C$ with $A$ on the ground. $A B$ is the tower of height $h$ m and $B C$ is the pole of 6 m on top of it. $P$ is a point on the ground with $A P = x$ m. Join $P B$ and $P C$ . Mark $∠ C P A = 6 0^{\circ}$ , and the angle of depression of $P$ from $B$ as $4 5^{\circ}$ , so $∠ B P A = 4 5^{\circ}$ [alternate angles]. $∠ B A P = 9 0^{\circ}$ .
1 mark
In right $△ B A P$ :
$tan 4 5^{\circ} = \frac{A B}{A P} \Rightarrow 1 = \frac{h}{x}$
1 mark
$\Rightarrow h = x$ ... (i)
½ mark
In right $△ C A P$ :
$tan 6 0^{\circ} = \frac{A C}{A P} \Rightarrow 3 = \frac{h + 6}{x}$
1 mark
$\Rightarrow h + 6 = 3 x$ ... (ii)
½ mark
Substitute (i) in (ii):
$h + 6 = 3 h \Rightarrow h = \frac{6}{3 - 1} = \frac{6 ( 3 + 1 )}{2} = 3 (3 + 1)$
$= 3 (2.73) = 8.19$ m
½ mark
$x = h = 8.19$ m.
Height of the tower $= 8.19$ m. Distance of P from the foot of the tower $= 8.19$ m.
½ mark

Answer: Height of tower = 3(√3+1) ≈ 8.19 m; distance of P from foot of tower ≈ 8.19 m

Where the marks are

Draws a correct figure with the tower, pole and both angles marked (1 mark)
Uses $tan 4 5^{\circ} = h / x$ (1 mark)
Gets $h = x$ (½ mark)
Uses $tan 6 0^{\circ} = (h + 6) / x$ (1 mark)
Gets $h + 6 = 3 x$ (½ mark)
Finds $h = 3 (3 + 1) \approx 8.19$ m (½ mark)
Finds distance $x \approx 8.19$ 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 paper2024 board paper, set 30/1/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2024: question 35

Hints

Step-by-step solution

Revise the chapter

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