Question : Two ships are on the opposite of a lighthouse such that all three of them are collinear. The angles of depression of the two ships from the top of the lighthouse are 30° and 60°. If the ships are $230 \sqrt{3}$ m apart, then find the height of the lighthouse (in m).
Option 1: 175.4
Option 2: 165.2
Option 3: 172.5
Option 4: 180.5
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Correct Answer: 172.5
Solution : Given: The distance between the ships is BC = $230\sqrt3$ m Let the height of the lighthouse AD be $h$ m. From $\triangle$ABD we get, $\tan30°=\frac{AD}{BD}$ ⇒ BD = $h\sqrt3$ Again from $\triangle$ACD we get, $\tan60°=\frac{AD}{CD}$ ⇒ CD = $\frac{h}{\sqrt3}$ According to the question, $h\sqrt3+\frac{h}{\sqrt3}=230\sqrt3$ ⇒ $3h+h=690$ ⇒ $4h=690$ ⇒ $h=\frac{690}{4}=172.5$ m Hence, the correct answer is 172.5 m.
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Question : Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from the two ships are 30° and 45°, respectively. If the lighthouse is 100 m high, the distance between the two ships is: (take $\sqrt{3}= 1.73$)
Option 1: 173 metres
Option 2: 200 metres
Option 3: 273 metres
Option 4: 300 metres
Question : The angle of depression of two ships from the top of a lighthouse is $60^{\circ}$ and $45^{\circ}$ towards the east. If the ships are 300 metres apart, the height of the lighthouse (in metres) is:
Option 1: $200\left ( 3+\sqrt{3} \right )$
Option 2: $250\left ( 3+{\sqrt3} \right )$
Option 3: $150\left (3 +{\sqrt3} \right)$
Option 4: $160 \left (3+{\sqrt3} \right)$
Question : A helicopter, at an altitude of 1500 metres, finds that two ships are sailing towards it, in the same direction. The angles of depression of the ships as observed from the helicopter are $60^{\circ}$ and $30^{\circ}$, respectively. Distance between the two ships, in metres, is:
Option 1: $1000\sqrt{3}$
Option 2: $\frac{1000}{{\sqrt3}}$
Option 3: $500{\sqrt3}$
Option 4: $\frac{500}{{\sqrt3}}$
Question : From the top of a cliff 100 metres high, the angles of depression of the top and bottom of a tower are 45° and 60°, respectively. The height of the tower is:
Option 1: $\frac{100}{3}(3-\sqrt{3})$ m
Option 2: $\frac{100}{3}(\sqrt3-1)$ m
Option 3: $\frac{100}{3}(2\sqrt3-1)$ m
Option 4: $\frac{100}{3}(\sqrt3-\sqrt{2})$ m
Question : A Navy captain is going away from a lighthouse at the speed of $4(\sqrt3–1)$ m/s. He observed that it took him 1 minute to change the angle of elevation of the top of the lighthouse from 60° to 45°. What is the height (in meters) of the lighthouse?
Option 1: $240\sqrt{3}$
Option 2: $480(\sqrt{3}–1)$
Option 3: $360\sqrt{3}$
Option 4: $280\sqrt{2}$
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