From a point on the ground, the angle of elevation of the top of a pedestal is 30° and that of the top of the flagstaff fixed on the pedestal is 60°. If the length of the flagstaff is 5 m, then find the height of the pedestal and its distance from the point of observation on ground. (Use √(3) = 1.73)

Height of pedestal = 2.5 m; distance 4.33 m

MathsCBSE Class 10 · Basic · Board exam 2025, set 430/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2025: question 35

Set in the CBSE Class 10 board exam of 2025, paper 430/4/1, section D, question 35, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q35 · 5 marks · Long answerSome Applications of Trigonometry

From a point on the ground, the angle of elevation of the top of a pedestal is $3 0^{\circ}$ and that of the top of the flagstaff fixed on the pedestal is $6 0^{\circ}$ . If the length of the flagstaff is 5 m, then find the height of the pedestal and its distance from the point of observation on ground. (Use $3 = 1.73$ )

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/4/1.

Hints

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

Hint 1

Draw the pedestal and the flagstaff as one vertical line, with the observer on the ground.

Hint 2

Write tan of each angle using the same horizontal distance $x$ .

Hint 3

Eliminate $x$ between the two equations to find $h$ first.

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 pedestal of height $h$ m, with foot B on the ground. AD is the flagstaff of 5 m on top, so BD = $(h + 5)$ m. C is the point of observation, with BC = $x$ m and $∠ A B C = 9 0^{\circ}$ . $∠ A C B = 3 0^{\circ}$ and $∠ D C B = 6 0^{\circ}$ .
1 mark
In right $Δ A B C$ : $tan 3 0^{\circ} = \frac{A B}{B C} \Rightarrow \frac{1}{3} = \frac{h}{x} \Rightarrow x = h 3$ ...(i)
1½ marks
In right $Δ D B C$ : $tan 6 0^{\circ} = \frac{D B}{B C} \Rightarrow 3 = \frac{h + 5}{x} \Rightarrow h + 5 = x 3$ ...(ii)
1½ marks
Substituting (i) in (ii): $h + 5 = h 3 \times 3 = 3 h \Rightarrow 2 h = 5 \Rightarrow h = 2.5$ m
½ mark
$x = 2.5 \times 1.73 = 4.325 \approx 4.33$ m. So the pedestal is 2.5 m high and the point of observation is 4.33 m from its foot.
½ mark

Answer: Height of pedestal $= 2.5$ m; distance $\approx 4.33$ m

Where the marks are

Correct figure (1 mark)
$tan 3 0^{\circ} = \frac{h}{x} \Rightarrow x = h 3$ (1½ marks)
$tan 6 0^{\circ} = \frac{h + 5}{x} \Rightarrow h + 5 = x 3$ (1½ marks)
$h = 2.5$ m (½ mark)
$x = 4.33$ m (½ mark)

Revise the chapter

Chapter: Some Applications of Trigonometry (notes, formulas, where marks are lost and every past question)
Topic: Line of Sight and Angle of Elevation

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 paper2025 board paper, set 430/4/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2025: question 35

Hints

Step-by-step solution

Revise the chapter

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