The angle of depression of the front end of a ship as seen from the top of 120 m high light house is 60 $^{\circ}$ . How far is the ship from the light house?

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Solution

$∠ P A B = ∠ A B C = 60^{\circ}$ (alternate angles)
$t a n 60^{\circ} = \frac{A C}{B C} \Rightarrow B C = \frac{A C}{t a n 60^{\circ}} = \frac{120}{\sqrt{3}} = 40 \sqrt{3} m ∴ t h e s l i p w a s a t a d i s t a n c e 40 \sqrt{3} m$

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The angle of depression of the front end of a ship as seen from the top of 120 m high light house is 60∘. How far is the ship from the light house?

∠PAB=∠ABC=60∘ (alternate angles) tan60∘=ACBC⇒BC=ACtan60∘=120√3=40√3 m ∴ the slip was at a distance 40√3 m

The angle of depression of the front end of a ship as seen from the top of 120 m high light house is 60 $^{\circ}$ . How far is the ship from the light house?

$∠ P A B = ∠ A B C = 60^{\circ}$ (alternate angles)
$t a n 60^{\circ} = \frac{A C}{B C} \Rightarrow B C = \frac{A C}{t a n 60^{\circ}} = \frac{120}{\sqrt{3}} = 40 \sqrt{3} m ∴ t h e s l i p w a s a t a d i s t a n c e 40 \sqrt{3} m$