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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 2 · Set 1 · Q37
Tejas is standing at the top of a building and observes a car at an angle of depression of 30° as it approaches the base of the building at a uniform…
Tejas is standing at the top of a building and observes a car at an angle of depression of $\displaystyle 30$° as it approaches the base of the building at a uniform speed. $\displaystyle 6$ seconds later, the angle of depression increases to $\displaystyle 60$°, and at that moment, the car is $\displaystyle 25$ m away from the building.
Based on the information given above, answer the following questions :
(i)
What is the height of the building ?
(ii)
What is the distance between the two positions of the car ?
(iii)
(a)
What would be the total time taken by the car to reach the foot of the building from the starting point?
OR
(b)
What is the distance of the observer from the car when it makes an angle of $\displaystyle 60$°?
Marking-scheme solution
(i)
In $\displaystyle \triangle \mathrm{ABC}$,
\[\begin{array}{l}
\tan 60^{\circ}=\sqrt{3}=\frac{\mathrm{AB}}{25} \\
\Rightarrow \mathrm{AB}=25 \sqrt{3}
\end{array}
\]
∴ Height of building $\displaystyle =25 \sqrt{3} \mathrm{~m}$
(ii)
In $\displaystyle \triangle \mathrm{ABD}$,
\[\begin{array}{c}
\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{25 \sqrt{3}}{\mathrm{BD}} \\
\Rightarrow \quad \mathrm{BD}=75
\end{array}
\]
∴ Distance between two positions of car $\displaystyle =75-25=50 \mathrm{~m}$
(iii)
Time taken to cover the distance of $\displaystyle 50 \mathrm{~m}=6 \mathrm{sec}$
\[\begin{aligned}
\therefore \text { Time taken to cover the distance of } 75 \mathrm{~m} & =\frac{6}{50} \times 75 \\
& =9 \mathrm{sec}
\end{aligned}
\]
In $\displaystyle \triangle \mathrm{ABC}$,
\[\begin{array}{r}
\cos 60^{\circ}=\frac{\mathrm{BC}}{\mathrm{AC}} \\
\Rightarrow \quad \frac{1}{2}=\frac{25}{\mathrm{AC}} \\
\Rightarrow \quad \mathrm{AC}=50
\end{array}
\]
∴ Distance of the observer from car when it makes the angle of $\displaystyle 60^{\circ}=50 \mathrm{~m}$
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