Passenger boarding stairs, sometimes referred to as boarding ramps, stair cars or aircraft steps, provide a mobile means to travel between the aircraft doors and the ground. Larger aircraft have door sills 5 to 20 feet (1 foot = 30 cm) high. Stairs facilitate safe boarding and de-boarding. An aircraft has a door sill at a height of 15 feet above the ground. A stair car is placed at a horizontal distance of 15 feet from the plane. Based on given information, answer the questions given in part (i) and (ii). (i) Find the angle at which stairs are inclined to reach the door sill 15 feet high above the ground. (ii) Find the length of stairs used to reach the door sill. (iii) (a) If the 20 feet long stairs is inclined at an angle of 60° to reach the door sill, then find the height of the door sill above the ground. (use √(3) = 1.732) (iii) (b) What should be the shortest possible length of stairs to reach the door sill of the plane 20 feet above the ground, if the angle of elevation cannot exceed 30°? Also, find the horizontal distance of base of stair car from the plane.

MathsCBSE Class 10 · Standard · Board exam 2025, set 30/5/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2025: question 38

Set in the CBSE Class 10 board exam of 2025, paper 30/5/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedSome Applications of Trigonometry

Passenger boarding stairs, sometimes referred to as boarding ramps, stair cars or aircraft steps, provide a mobile means to travel between the aircraft doors and the ground. Larger aircraft have door sills 5 to 20 feet (1 foot = 30 cm) high. Stairs facilitate safe boarding and de-boarding.

An aircraft has a door sill at a height of 15 feet above the ground. A stair car is placed at a horizontal distance of 15 feet from the plane.

Based on given information, answer the questions given in part (i) and (ii).

(i)
Find the angle at which stairs are inclined to reach the door sill 15 feet high above the ground.
1 mark
(ii)
Find the length of stairs used to reach the door sill.
1 mark
(iii)
(a) If the 20 feet long stairs is inclined at an angle of $6 0^{\circ}$ to reach the door sill, then find the height of the door sill above the ground. (use $3 = 1.732$ )
2 marks
OR
(iii)
(b) What should be the shortest possible length of stairs to reach the door sill of the plane 20 feet above the ground, if the angle of elevation cannot exceed $3 0^{\circ}$ ? Also, find the horizontal distance of base of stair car from the plane.
2 marks

Figure in the paper: Left: an aircraft with its front door open, a passenger with luggage walking down a set of boarding stairs with railings to the tarmac. Right: a line drawing of a boarding stair car labelled 'STAIRS', with a sloped staircase with railings and a base platform on wheels at ground level. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/5/1.

Hints

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

Hint 1

Sketch a right triangle: the door height is the vertical side, the ground distance is the base, and the stairs are the hypotenuse.

Hint 2

Pick the ratio that links what you know to what you need: tan for height and base, sin for height and hypotenuse.

Hint 3

In (iii)(b), a steeper stair is shorter, so use the largest allowed angle, $3 0^{\circ}$ .

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.

(i) Draw a right triangle: vertical side (door sill height) 15 ft, horizontal side 15 ft, stairs as hypotenuse $l$ , with inclination $θ$ at the ground. Then $tan θ = \frac{15}{15} = 1$
½ mark
(i) $\Rightarrow θ = 4 5^{\circ}$
½ mark
(ii) $sin 4 5^{\circ} = \frac{15}{l} \Rightarrow \frac{1}{2} = \frac{15}{l}$
½ mark
(ii) $l = 152 \approx 21.21$ ft
½ mark
(iii)(a) The stairs (20 ft) form the hypotenuse and $h$ is the height. $sin 6 0^{\circ} = \frac{h}{20} = \frac{3}{2}$
1 mark
(iii)(a) $h = 103$
½ mark
(iii)(a) $h = 10 \times 1.732 = 17.32$ ft
½ mark
(iii)(b) The height is 20 ft. For a fixed height, the stair length $l = \frac{20}{sin θ}$ is shortest when $θ$ is largest, i.e. $θ = 3 0^{\circ}$ . Then $sin 3 0^{\circ} = \frac{20}{l} = \frac{1}{2}$
½ mark
(iii)(b) $l = 40$ ft
½ mark
(iii)(b) Let $x$ be the horizontal distance. $tan 3 0^{\circ} = \frac{20}{x} = \frac{1}{3}$
½ mark
(iii)(b) $x = 203 \approx 34.64$ ft
½ mark

Answer: (i) 45°; (ii) $152$ ft ≈ 21.21 ft; (iii)(a) 17.32 ft; (iii)(b) shortest length 40 ft, horizontal distance $203$ ft ≈ 34.64 ft

Where the marks are

(i) $tan θ = 1$ (½ mark); $θ = 4 5^{\circ}$ (½ mark)
(ii) Uses $sin 4 5^{\circ} = 15/ l$ (½ mark); $l = 152$ ft (½ mark)
(iii)(a) $sin 6 0^{\circ} = h /20$ (1 mark); $h = 103$ (½ mark); 17.32 ft (½ mark)
(iii)(b) $sin 3 0^{\circ} = 20/ l$ (½ mark); $l = 40$ ft (½ mark); $tan 3 0^{\circ} = 20/ x$ (½ mark); $x = 203$ ft (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 30/5/1 (Standard): 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 Standard 2025: question 38

Hints

Step-by-step solution

Revise the chapter

More questions from Some Applications of Trigonometry