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Q:

The height of a light house is 20 mts above sea level. The angle of depression (from the top of the lighthouse) of a ship in the sea is 30 deg. What is the distance of the ship from the foot of the light house?

A) 16 m	B) 20√3 m
C) 20 m	D) 30 m

Answer & Explanation Answer: B) 20√3 m

Explanation:

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Filed Under: Height and Distance
Exam Prep: Bank Exams

Q:

A Navy captain going away from a lighthouse at the speed of 4[(√3) – 1] m/s. He observes that it takes him 1 minute to change the angle of elevation of the top of the lighthouse from 60 to 45 deg. What is the height (in metres) of the lighthouse?

A) 240√3	B) 480[(√3) – 1]
C) 360√3	D) 280√2

Answer & Explanation Answer: A) 240√3

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Q:

A boat is sailing towards a lighthouse of height 20√3 m at a certain speed. The angle of elevation of the top of the lighthouse changes from 30° to 60° in 10 seconds. What is the time taken (in seconds) by the boat to reach the lighthouse from its initial position?

A) 10	B) 15
C) 20	D) 60

Answer & Explanation Answer: B) 15

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Filed Under: Quantitative Aptitude - Arithmetic Ability
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Q:

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 30º and 45º 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

Answer & Explanation Answer: C) 273 m

Explanation:

Let AB be the lighthouse and C and D be the positions of the ships.

Then, AB = 100m, $∠ A C B = 30 ° a n d ∠ A D B = 45 °$

$\frac{A B}{A C} = \tan 30 ° = \frac{1}{\sqrt{3}} = > A C = A B * \sqrt{3} = 100 \sqrt{3 m}$

$\frac{A B}{A D} = \tan 45 ° = 1 = > A D = A B = 100 m$

CD=(AC+AD)= $(100 \sqrt{3} + 100) m = 100 (\sqrt{3} + 1) = 100 * 2.73 = 273 m$

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Filed Under: Height and Distance

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