As observed from the top of a $75 m$ high lighthouse from the sea-level, the angle of depression of two ships are $30$ and $45$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

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Solution

Let the ships be A and B and OH be the light house
$∴ tan 45 ° = \frac{O H}{O A}$
$O A = O H$
$O A = 75 m$
$∴ tan 30 ° = \frac{O H}{O B}$
$\frac{1}{\sqrt{3}} = \frac{O H}{O B}$
$O B = O H \sqrt{3}$
$= 75 \sqrt{3} m$
Distance between ships is AB
$∴ A B = O B - A B$
$= 75 \sqrt{3} - 75 m$
$= 75 (\sqrt{3} - 1) m$
$A B = 54.90 m$

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As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are $30^{\circ}$ and $45^{\circ}$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Q. As observed from the top of a

75 m

high lighthouse from the sea-level, the angles of depression of two ships arc

30^{o}

and

45^{o}

. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

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As observed from the top of a 75m high lighthouse from the sea-level, the angle of depression of two ships are 30 and 45. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Let the ships be A and B and OH be the light house∴tan45°=OHOAOA=OHOA=75m∴tan30°=OHOB1√3=OHOBOB=OH√3=75√3mDistance between ships is AB∴AB=OB−AB=75√3−75m=75(√3−1)mAB=54.90m

As observed from the top of a $75 m$ high lighthouse from the sea-level, the angle of depression of two ships are $30$ and $45$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.