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Mathematics · 2025 · 5 marks
CBSE 2025 · Region 2 · Set 1 · Q32
(a) Two ships are sailing in the sea on either side of a lighthouse. The angles of depression to the two ships as observed from the top of the…
Two ships are sailing in the sea on either side of a lighthouse. The angles of depression to the two ships as observed from the top of the lighthouse are $\displaystyle 60$° and $\displaystyle 45$°, respectively. If the distance between the ships is $\displaystyle 100\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right) \mathrm{m}$, then find the height of the lighthouse.
The angles of depression of the top and the bottom of an $\displaystyle 8$ m tall building from the top of another multistoried building are $\displaystyle 30$° and $\displaystyle 45$°, respectively. Find the height of the multistoried building and the distance between the two buildings.
Marking-scheme solution
Here, AB represents the height of the lighthouse.
In right \(\displaystyle \Delta \mathrm{ABP}\)
\[\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{PB}}=\tan 60^{\circ}=\sqrt{3} \\
& \Rightarrow \mathrm{PB}=\frac{\mathrm{AB}}{\sqrt{3}} \quad ---(1)
\end{aligned}
\]
In right \(\displaystyle \Delta \mathrm{ABQ}\)
\[\begin{aligned}
& \frac{\mathrm{AB}}{\mathrm{BQ}}=\tan 45^{\circ}=1 \\
& \Rightarrow \mathrm{BQ}=\mathrm{AB} \quad ---(2)
\end{aligned}
\]
Adding ($\displaystyle 1$) and ($\displaystyle 2$), we have
\[\begin{aligned}
& \mathrm{PB}+\mathrm{BQ}=\frac{\mathrm{AB}}{\sqrt{3}}+\mathrm{AB} \\
& \Rightarrow \mathrm{PQ}=\mathrm{AB}\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right) \\
& \Rightarrow 100\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)=\mathrm{AB}\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right) \\
& \Rightarrow \mathrm{AB}=100 \mathrm{~m}
\end{aligned}
\]
Here, CD represents the multistoried building.
In right \(\displaystyle \Delta \mathrm{DEA}\)
\[\begin{aligned}
& \frac{\mathrm{DE}}{\mathrm{AE}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \mathrm{AE}=\sqrt{3} \mathrm{DE} \quad ---(1)
\end{aligned}
\]
In right \(\displaystyle \Delta \mathrm{DCB}\)
\[\begin{aligned}
& \frac{\mathrm{DC}}{\mathrm{BC}}=\tan 45^{\circ}=1 \\
& \Rightarrow \mathrm{BC}=\mathrm{DC} \quad ---(2)
\end{aligned}
\]
From figure, \(\displaystyle \mathrm{AE}=\mathrm{BC}\)
\[\begin{aligned}
& \therefore \sqrt{3} \mathrm{DE}=\mathrm{DC} \\
& \Rightarrow \sqrt{3}(\mathrm{DC}-8)=\mathrm{DC} \\
& \Rightarrow \mathrm{DC}=\frac{8 \sqrt{3}}{\sqrt{3}-1} \\
& \quad=\frac{8 \sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}=(12+4 \sqrt{3}) \mathrm{m}
\end{aligned}
\]
From ($\displaystyle 2$), \(\displaystyle \mathrm{BC}=(12+4 \sqrt{3}) \mathrm{m}\)
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