From the top of a lighthouse, an observer looks at a ship and finds the angle of depression to be $60^{o}$ . If the height of the lighthouse is $93$ metres, then find how far is that ship from the lighthouse?

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Solution

Given that height of the light house is $93$ meters
Given that the angle of depression is $60^{0}$
Let the distance of the ship from the foot of light house be $x$
So we have $t a n (30^{0}) = \frac{x}{93}$
$\Rightarrow x = 93 t a n (30^{0}) = \frac{93}{\sqrt{3}} = 53.69$

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From the top of a lighthouse, an observer looks at a ship and finds the angle of depression to be 60o. If the height of the lighthouse is 93 metres, then find how far is that ship from the lighthouse?

Given that height of the light house is 93 metersGiven that the angle of depression is 600Let the distance of the ship from the foot of light house be xSo we have tan(300)=x93⇒x=93tan(300)=93√3=53.69

From the top of a lighthouse, an observer looks at a ship and finds the angle of depression to be $60^{o}$ . If the height of the lighthouse is $93$ metres, then find how far is that ship from the lighthouse?

Given that height of the light house is $93$ meters
Given that the angle of depression is $60^{0}$
Let the distance of the ship from the foot of light house be $x$
So we have $t a n (30^{0}) = \frac{x}{93}$
$\Rightarrow x = 93 t a n (30^{0}) = \frac{93}{\sqrt{3}} = 53.69$