CBSE Class 10 Mathematics Standard 2023: question 33(a)

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Step-by-step solution

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1. Figure: PQ is the lighthouse with $PQ=75$ m and foot P at sea level. QX is the horizontal through Q. R and S are the ships on the same side, in line with P, with R nearer. The angles of depression are $\angle XQR=60^{\circ }$ and $\angle XQS=30^{\circ }$. As alternate angles, $\angle QRP=60^{\circ }$ and $\angle QSP=30^{\circ }$.

   1 mark

2. In right $△QPR$: $\tan 60^{\circ }=\frac{PQ}{PR}\Rightarrow \sqrt{3}=\frac{75}{PR}\Rightarrow PR=\frac{75}{\sqrt{3}}=25\sqrt{3}$ m

   1½ marks

3. In right $△QPS$: $\tan 30^{\circ }=\frac{PQ}{PS}\Rightarrow \frac{1}{\sqrt{3}}=\frac{75}{PS}\Rightarrow PS=75\sqrt{3}$ m

   1 mark

4. Distance between the ships: $RS=PS-PR=75\sqrt{3}-25\sqrt{3}=50\sqrt{3}$ m

   1 mark

5. $=50\times 1.73=86.5$ m

   ½ mark

Answer: Distance between the ships = $50\sqrt{3}$ m = 86.5 m

Revise the chapter

• Chapter: Some Applications of Trigonometry (notes, formulas, where marks are lost and every past question)
• Topic: Angle of Depression