# 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:

A. 173 m

B. 200 m

C. 273 m

D. 300 m

E. None of these

**Answer: Option C **

__Solution(By Examveda Team)__

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**

Related Questions on Height and Distance

A. 173 m

B. 200 m

C. 273 m

D. 300 m

E. None of these

A. 4 √3 units

B. 8 units

C. 12 units

D. Data inadequate

E. None of these

A. 21.6 m

B. 23.2 m

C. 24.72 m

D. None of these

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