Two ships are sailling in the sea on the two sides of a lighthouse. If the distance between the ships is 10 ( 3 + 1) meters and their angle of elevations of the top of the lighthouse are 60 ^ o ~and& 45 ^ o , then the

Correct answer: C. 10 3 meters Let,& AC is the lighthouse and the ships are at B and D . In Δ ABC, (AC)/(BC) = tan 60 ^ o where, AC = h meters = height of lighthouse (h)/(BC) = 3 ⇒ BC = h 3 meters In Δ ACD, (AC)/(CD) = tan 45 ^ o ⇒ (h)/(CD) = 1 ⇒ CD = h BC + CD = h 3 + h = 10 ( 3 + 1) ⇒ h = 10 3 meters

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Heights and Distances

NTA Abhyas 2020 · Mathematics · Heights and Distances

Two ships are sailling in the sea on the two sides of a lighthouse....

Two ships are sailling in the sea on the two sides of a lighthouse. If the distance between the ships is

10

\left(\sqrt{3} +

1\right)

meters and their angle of elevations of the top of the lighthouse are

{60}^{o}

~and&

{45}^{o}

, then the height of the lighthouse is (The two ships and the foot of lighthouse are in a straight line) Two ships are sailling in the sea on the two sides of a lighthouse. If the distance between the ships is 10 ( 3 + 1)

Two ships are sailling in the sea on the two sides of a lighthouse. If the distance between the ships is 10 ( 3 + 1)

Options

A $20$ meters
B $20\sqrt{3}$ meters
C $10\sqrt{3}$ meters
D $5\sqrt{3}$ meters

Correct answer

C. $10\sqrt{3}$ meters

Step-by-step solution

Let,&

AC

is the lighthouse and the ships are at

B

and

D .

\Delta ABC,

\dfrac{AC}{BC} = tan {60}^{o}

{where,

AC = h

meters = height of lighthouse}

\dfrac{h}{BC} = \sqrt{3} ⇒ BC = \dfrac{h}{\sqrt{3}}

meters
In

\Delta ACD,

\dfrac{AC}{CD} = tan {45}^{o} ⇒ \dfrac{h}{CD} = 1 ⇒ CD = h

BC + CD = \dfrac{h}{\sqrt{3}} + h = 10 \left(\sqrt{3} +

1\right)

<br />

⇒ h =

10\sqrt{3}meters

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