(a) As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 60°. If one ship is…

\(\displaystyle \mathrm{PQ}=\) 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 \(\displaystyle \triangle \mathrm{APC}, \tan 60^{\circ}=\frac{\mathrm{AP}}{\mathrm{AC}}\)

\[\begin{aligned} & \sqrt{3}=\frac{30+\mathrm{h}}{30 \sqrt{3}} \Rightarrow 30 \sqrt{3} \times \sqrt{3}=30+\mathrm{h} \\ & \Rightarrow \mathrm{~h}=30(3-1) \\ & \Rightarrow \mathrm{h}=60 \\ & \therefore \text { Height of transmission tower }=60 \mathrm{~m} \end{aligned} \]