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Mathematics · 2025 · 4 marks
CBSE 2025 · Region 5 · Set 1 · Q38
Passenger boarding stairs, sometimes referred to as boarding ramps, stair cars or aircraft steps, provide a mobile means to travel between the…
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 $\displaystyle 5$ to $\displaystyle 20$ feet ($\displaystyle 1$ foot = $\displaystyle 30$ cm) high. Stairs facilitate safe boarding and de-boarding.
An aircraft has a door sill at a height of $\displaystyle 15$ feet above the ground. A stair car is placed at a horizontal distance of $\displaystyle 15$ 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 $\displaystyle 15$ feet high above the ground.
(ii)
Find the length of stairs used to reach the door sill. Further, answer any one of the following questions :
(iii)
(a)
If the $\displaystyle 20$ feet long stairs is inclined at an angle of $\displaystyle 60$° to reach the door sill, then find the height of the door sill above the ground. (use $\displaystyle \sqrt{3}=1.732$ )
OR
(b)
What should be the shortest possible length of stairs to reach the door sill of the plane $\displaystyle 20$ feet above the ground, if the angle of elevation cannot exceed $\displaystyle 30$° ? Also, find the horizontal distance of base of stair car from the plane.
Marking-scheme solution
(i)
\(\displaystyle \tan \theta=\frac{15}{15}=1\)
\(\displaystyle \Rightarrow \theta=45^{\circ}\)
(ii)
\(\displaystyle \frac{15}{l}=\sin 45^{\circ}\)
\(\displaystyle \Rightarrow l=15 \sqrt{2} \mathrm{~ft} .\) or \(\displaystyle 21.21 \mathrm{~ft} .\) approx.
(iii)
\[\begin{aligned}
\frac{\mathrm{h}}{20} & =\sin 60^{\circ}=\frac{\sqrt{3}}{2} \\
\Rightarrow \mathrm{~h} & =10 \sqrt{3} \\
& =10 \times 1.732 \\
& =17.32 \mathrm{ft} .
\end{aligned}
\]
\(\displaystyle \frac{20}{l}=\sin 30^{\circ}=\frac{1}{2}\)
\(\displaystyle \Rightarrow l=40 \mathrm{~ft} .\)
\(\displaystyle \frac{20}{\mathrm{x}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)
\(\displaystyle \Rightarrow \mathrm{x}=20 \sqrt{3} \mathrm{~ft} .\) or \(\displaystyle 34.64 \mathrm{~ft} .\) approx.
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CBSE Class 10 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.