Question

from the top of 60m high light house, the angle of depression of the ship is 60°.find the distance between the ship and the foot of the light house.

Answer 1

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һєяє' ʏȏȗя ѧṅśwєя ʟȏȏҡıṅɢ ғȏя________
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✯♦let AB the height of the light house

♦BC and BD be the distances of ships

therefore,

▶/_ACB = 45°

AB = BC = 60 m

♦since , /_ADB = 30°

▶therefore ,

$tan \: 30° = \frac{AB}{BD}$

$\sqrt[1]{3} = \frac{60} {60} + CD$

♦60 + CD = 60 × 1.732

♦CD= 103 .92 - 60

$= 43.92m$

$so \: the \: distance \: is \: 43.92$

$hope \: it \: helps$

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★BRAINLY★

Answer 2

Hey...


Here is your Answer

⭕If AB the heigth of the house...
And
⭕BC and BD the distance of ships....

Then..

Angle ABC = 45 DEGREE

AB = BC = 60 m

Then..

Angle ADB will be 30 degree

So..

Tan 30A degree will be equal to AB upon BD

⭕ 60 + CD =60 × 1.732

Then..

⭕ CD = 103.92- 60

43.92m will be distance...

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I HOPE IT WILL HELP U SIS✌✌✌✌