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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 2 · Set 1 · Q33
As observed from the top of a $\displaystyle 75$ m high lighthouse from the sea-level, the angles of depression of two ships are $\displaystyle 30$° and $\displaystyle 60$°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. (Use $\displaystyle \sqrt{3}=1 \cdot 73$ )From a point on the ground, the angle of elevation of the bottom and top of a transmission tower fixed at the top of $\displaystyle 30$ m high building are $\displaystyle 30$° and $\displaystyle 60$°, respectively. Find the height of the transmission tower. (Use $\displaystyle \sqrt{3}=1 \cdot 73$ )
As observed from the top of a $\displaystyle 75$ m high lighthouse from the sea-level, the angles of depression of two ships are $\displaystyle 30$° and $\displaystyle 60$°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. (Use $\displaystyle \sqrt{3}=1 \cdot 73$ )
From a point on the ground, the angle of elevation of the bottom and top of a transmission tower fixed at the top of $\displaystyle 30$ m high building are $\displaystyle 30$° and $\displaystyle 60$°, respectively. Find the height of the transmission tower. (Use $\displaystyle \sqrt{3}=1 \cdot 73$ )
Marking-scheme solution
\(\displaystyle \mathrm{PQ}=\) Height of Light house \(\displaystyle =75 \mathrm{~m}\)
\[\begin{aligned}
& \angle \mathrm{XQS}=\angle \mathrm{QSP}=30^{\circ} \\
& \angle \mathrm{XQR}=\angle \mathrm{QRP}=60^{\circ}
\end{aligned}
\]
R and S are position of ships.
In \(\displaystyle \Delta \mathrm{PQR}\),
\[\frac{75}{\mathrm{PR}}=\tan 60^{\circ}=\sqrt{3} \Rightarrow \mathrm{PR}=\frac{75}{\sqrt{3}}=25 \sqrt{3}
\]
In \(\displaystyle \Delta \mathrm{PQS}, \frac{75}{\mathrm{PS}}=\tan 30^{\circ}\)
\[\Rightarrow \mathrm{PS}=75 \sqrt{3}
\]
∴ Distance between the ships, \(\displaystyle \mathrm{RS}=\mathrm{PS}-\mathrm{PR}\)
\[\begin{aligned}
& =75 \sqrt{3}-25 \sqrt{3}=50 \sqrt{3} \\
& =50 \times 1 \cdot 73=86 \cdot 5
\end{aligned}
\]
∴ Distance between the ships is $\displaystyle 86.5$ m
Height of building \(\displaystyle \mathrm{AB}=30 \mathrm{~m}\) BP = transmission tower = h(say) \(\displaystyle \angle \mathrm{ACB}=30^{\circ}, \angle \mathrm{ACP}=60^{\circ}\) In \(\displaystyle \triangle \mathrm{ABC}, \tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AC}} \left.\Rightarrow \frac{1}{\sqrt{3}}=\frac{30}{\mathrm{AC}} \Rightarrow \mathrm{AC}=30 \sqrt{3}\right]\) In \(\displaystyle \triangle \mathrm{APC}, \tan 60^{\circ}=\frac{\mathrm{AP}}{\mathrm{AC}}\)
\[\begin{aligned}
& \sqrt{3}=\frac{30+\mathrm{h}}{30 \sqrt{3}} \Rightarrow 30 \sqrt{3} \times \sqrt{3}=30+\mathrm{h} \\
& \Rightarrow \mathrm{~h}=30(3-1) \\
& \Rightarrow \mathrm{h}=60 \\
& \therefore \text { Height of transmission tower }=60 \mathrm{~m}
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.