Anil looked up at the top of a lighthouse from his boat, and found the angle of elevation to be 30 $^{\circ}$ . After sailing in a straight line 50 m towards the lighthouse, he found that the angle of elevation changed to 45 $^{\circ}$ . Find the height of the lighthouse.

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25√3

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25(√3-1)

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25(√3+1)

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Solution

The correct option is D
25(√3+1)

Now since tan 45 = 1,BC=DB=x
Also, tan 30 = $\frac{1}{\sqrt{3}} = \frac{x}{50 + x}$
Thus x( $\sqrt{3}$ - 1) = 50
or, x = 25( $\sqrt{3}$ + 1)m

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Similar questions

Two boats approached a lighthouse in the midsea from opposite directions the angle of elevation of the top of the lighthouse are 30 and 45 degree respectively if the distance between two boats is 100m find the highest of the lighthouse

A man rowing a boat away from a lighthouse which is 100m high takes 2 mins to change the angle of elevation of the lighthouse from $60^{\circ} t o 45^{\circ}$ . The speed of the boat is:

Q. A man rowing a boat away from a lighthouse which is 100 m high takes 2 min to change the angle of elevation of the light house from

45^{\circ}

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. Find the speed of the boat.

[T a n 45^{\circ} = 1, T a n 30^{\circ} = 0.57]

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Anil looked up at the top of a lighthouse from his boat, and found the angle of elevation to be 30∘. After sailing in a straight line 50 m towards the lighthouse, he found that the angle of elevation changed to 45∘. Find the height of the lighthouse.

The correct option is D 25(√3+1) Now since tan 45 = 1,BC=DB=x Also, tan 30 = 1√3=x50+x Thus x(√3 - 1) = 50 or, x = 25(√3 + 1)m

Anil looked up at the top of a lighthouse from his boat, and found the angle of elevation to be 30 $^{\circ}$ . After sailing in a straight line 50 m towards the lighthouse, he found that the angle of elevation changed to 45 $^{\circ}$ . Find the height of the lighthouse.

The correct option is D
25(√3+1)

Now since tan 45 = 1,BC=DB=x
Also, tan 30 = $\frac{1}{\sqrt{3}} = \frac{x}{50 + x}$
Thus x( $\sqrt{3}$ - 1) = 50
or, x = 25( $\sqrt{3}$ + 1)m