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Mathematics · 2025 · 4 marks
CBSE 2025 · Region 1 · Set 1 · Q38
Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be 60^°.…
Amrita stood near the base of a lighthouse, gazing up at its towering height. She measured the angle of elevation to the top and found it to be $\displaystyle 60^{\circ}$. Then, she climbed a nearby observation deck, $\displaystyle 40$ metres higher than her original position and noticed the angle of elevation to the top of lighthouse to be $\displaystyle 45$°.
Based on the above given information, answer the following questions :
(i)
If CD is h metres, find the distance BD in terms of 'h'.
(ii)
Find distance BC in terms of ' h '.
(iii)
(a)
Find the height CE of the lighthouse [Use $\displaystyle \sqrt{3}=1 \cdot 73$ ]
OR
(b)
Find distance AE , if $\displaystyle \mathrm{AC}=100 \mathrm{~m}$.
Marking-scheme solution
(i) \(\displaystyle \frac{\mathrm{h}}{\mathrm{BD}}=\tan 45^{\circ}=1\)
\[\Rightarrow \mathrm{BD}=\mathrm{h} \mathrm{~m}
\]
(ii) \(\displaystyle \frac{\mathrm{h}}{\mathrm{BC}}=\sin 45^{\circ}=\frac{1}{\sqrt{2}}\)
\[\Rightarrow \mathrm{BC}=\sqrt{2} \mathrm{~h} \mathrm{~m}
\]
(iii) (a) \(\displaystyle \tan 60^{\circ}=\frac{\mathrm{EC}}{\mathrm{AE}}\)
\[\begin{aligned}
& \Rightarrow \sqrt{3}=\frac{\mathrm{h}+40}{\mathrm{~h}} \\
& \Rightarrow \quad \mathrm{~h}=20(\sqrt{3}+1)=20 \times 2.73=54.6 \mathrm{~m} \\
& \therefore \mathrm{CE}=54.6+40=94.6 \mathrm{~m}
\end{aligned}
\]
(iii)
\(\displaystyle \cos 60^{\circ}=\frac{\mathrm{AE}}{\mathrm{AC}}\)
\[\Rightarrow \frac{1}{2}=\frac{\mathrm{AE}}{100}
\]
\[\therefore \quad \mathrm{AE}=50 \mathrm{~m}
\]
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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.