Question

From the top of a light house ,it is observed that a ship is sailing directly towards it and the angle of depression of the ship changez from 30 to 45 in 10 min Assuming that the ship is sailling with uniform speed .calculate in how much more time will the ship reach the light house

Answer 1

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Answer 2

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Step-by-step explanation:

Refer the attached figure

Let the height of the lighthouse be H

The angle of depression of the ship changes from 30 to 45 in 10 min

In ΔABD

$tan \theta = \frac{perpendicular}{Base}$

$tan30^{\circ} = \frac{AB}{BD}$

$\frac{1}{\sqrt{3}} = \frac{H}{BD}$

$BD= \frac{H}{\frac{1}{\sqrt{3}} }$

In ΔABC

$tan \theta = \frac{perpendicular}{Base}$

$tan45^{\circ} = \frac{AB}{BC}$

$1= \frac{H}{BC}$

$BC= H$

Now distance traveled in 10 minutes = BD -BC = $H\sqrt{3}-H$

$Speed = \frac{Distance}{Time}$

$Speed = \frac{H\sqrt{3}-H}{10}$

If the ship reaches the light house the distance becomes 0 from H

So, Time = $\frac{Distance}{Speed}$

Time = $\frac{H-0}{\frac{H\sqrt{3}-H}{10}}$

Time = $13.66$

So, The ship will reach the light house in 13.66 minutes