A man is standing on the deck of a ship, which is $40 \; m$ above water level. He observes the angle of elevation of the top of a hill as $60^\circ$ and the angle of depression of the base of the hill as $30^\circ$. Calculate the distance of the hill from the ship and the height of the hill. [Given: $\sqrt{3} = 1.732$]
$M =$ Position of man on the deck of a ship
$O = $ Position of the ship
$MO = 40 \; m$ Height of the deck of ship above water level
$H =$ Position of the hill
$HH' =$ Height of the hill
$OH =$ Distance of the hill from the ship
Draw $MP \perp HH'$
Then, $\;$ $MP = OH$, $MO = PH = 40 \; m$
In $\triangle OMH$, $\;$ $\dfrac{OM}{OH} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$
$\implies$ $OH = OM \times \sqrt{3} = 40 \times \sqrt{3} = 40 \times 1.732 = 69.28$
In $\triangle MPH'$, $\;$ $\dfrac{H'P}{MP} = \tan 60^\circ = \sqrt{3}$
$\implies$ $H'P = MP \times \sqrt{3} = 40 \sqrt{3} \times \sqrt{3} = 120$
Now, $HH' = H'P + PH = 120 + 40 = 160$
$\therefore \;$ Distance of the hill from the ship $= 69.28 \; m$
Height of hill $= 160 \; m$