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Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30° and 45° respectively. If the lighthouse is 100 m high, the distance between the two ships is: ages is:

a) 173 m
b) 200 m
c) 273 m
d) 300 m

Answer: c
Explanation: Let AB be the lighthouse and C and D be the positions of the ships.
Then, AB = 100m, ∠ACB = 30° and ∠ADB = 45°
$$\frac{{AB}}{{AC}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }}$$
$$ \Rightarrow AC = AB \times \sqrt 3 =100 \sqrt 3{\text{m}}$$
$$\frac{{AB}}{{AD}} = \tan {45^ \circ } = 1$$
$$ \Rightarrow AD = AB = 100{\text{m}}$$
CD = (AC+AD) = (100√3 +100)
= 100(√3 +1) = 100(1.73+1) =100 × 2.73 = 273m

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