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A lighthouse stands tall on a cliff by the sea, watching over ships that pass by. One day a ship is seen approaching the shore and from the top of the lighthouse, the angles of depression of the ship are observed to be $30^{\circ}$ and $45^{\circ}$ as it moves from point P to point Q. The height of the lighthouse is $50$ metres.
Based on the information given above, answer the following questions:
(i) Find the distance of the ship from the base of the lighthouse when it is at point Q, where the angle of depression is $45^{\circ}$.
(ii) Find the measures of $\angle PBA$ and $\angle QBA$.
(iii) (a) Find the distance travelled by the ship between points P and Q.
OR
(b) If the ship continues moving towards the shore and takes $10$ minutes to travel from Q to A, calculate the speed of the ship in $\operatorname{km/h}$, from Q to A.
Based on the information given above, answer the following questions:
(i) Find the distance of the ship from the base of the lighthouse when it is at point Q, where the angle of depression is $45^{\circ}$.
(ii) Find the measures of $\angle PBA$ and $\angle QBA$.
(iii) (a) Find the distance travelled by the ship between points P and Q.
OR
(b) If the ship continues moving towards the shore and takes $10$ minutes to travel from Q to A, calculate the speed of the ship in $\operatorname{km/h}$, from Q to A.
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Sol. (i) $\angle AQB = \angle QBX = 45^{\circ}$ and $\angle APB = \angle PBX = 30^{\circ}$
In $\triangle AQB$, $\tan 45^{\circ} = \frac{50}{AQ}$
$AQ = 50 \operatorname{m}$
(ii) $\angle PBA = 60^{\circ}$
$\angle QBA = 45^{\circ}$
(iii)(a) In $\triangle APB$, $\tan 30^{\circ} = \frac{50}{AP}$
$AP = 50\sqrt{3} \operatorname{m}$
Distance travelled by the ship $= PQ = 50\sqrt{3} - 50 = 50(\sqrt{3} - 1) \operatorname{m}$
or $36.5 \operatorname{m}$
OR
(iii)(b) Speed of the ship $= \frac{50 \operatorname{metres}}{10 \operatorname{minutes}}$
$= 0.3 \operatorname{km/h}$
In $\triangle AQB$, $\tan 45^{\circ} = \frac{50}{AQ}$
$AQ = 50 \operatorname{m}$
(ii) $\angle PBA = 60^{\circ}$
$\angle QBA = 45^{\circ}$
(iii)(a) In $\triangle APB$, $\tan 30^{\circ} = \frac{50}{AP}$
$AP = 50\sqrt{3} \operatorname{m}$
Distance travelled by the ship $= PQ = 50\sqrt{3} - 50 = 50(\sqrt{3} - 1) \operatorname{m}$
or $36.5 \operatorname{m}$
OR
(iii)(b) Speed of the ship $= \frac{50 \operatorname{metres}}{10 \operatorname{minutes}}$
$= 0.3 \operatorname{km/h}$