CBSE 2025 · Region 3 · Set 1 · Q37 · 4 marks
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 $\displaystyle 30$° and $\displaystyle 45^{\circ}$ as it moves from point P to point Q. The height of the lighthouse is $\displaystyle 50$ 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 $\displaystyle 45^{\circ}$.(ii)Find the measures of $\displaystyle \angle \mathrm{PBA}$ and $\displaystyle \angle \mathrm{QBA}$.(iii)Find the distance travelled by the ship between points P and Q.If the ship continues moving towards the shore and takes $\displaystyle 10$ minutes to travel from Q to A, calculate the speed of the ship in km/h, from Q to A.
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 $\displaystyle 30$° and $\displaystyle 45^{\circ}$ as it moves from point P to point Q. The height of the lighthouse is $\displaystyle 50$ 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 $\displaystyle 45^{\circ}$.
(ii)
Find the measures of $\displaystyle \angle \mathrm{PBA}$ and $\displaystyle \angle \mathrm{QBA}$.
(iii)
Find the distance travelled by the ship between points P and Q.
If the ship continues moving towards the shore and takes $\displaystyle 10$ minutes to travel from Q to A, calculate the speed of the ship in km/h, from Q to A.
Marking-scheme solution
(i)
\(\displaystyle \angle \mathrm{AQB}=\angle \mathrm{QBX}=45^{\circ}\) and \(\displaystyle \angle \mathrm{APB}=\angle \mathrm{PBX}=30^{\circ}\)
In \(\displaystyle \triangle \mathrm{AQB}, \tan 45^{\circ}=\frac{50}{\mathrm{AQ}}\)
(ii)
\(\displaystyle \angle \mathrm{PBA}=60^{\circ}\)
\(\displaystyle \angle \mathrm{QBA}=45^{\circ}\)
(a)
In \(\displaystyle \triangle \mathrm{APB}, \tan 30^{\circ}=\frac{50}{\mathrm{AP}}\)
\(\displaystyle \mathrm{AP}=50 \sqrt{3} \mathrm{~m}\)
Distance travelled by the ship \(\displaystyle =\mathrm{PQ}=50 \sqrt{3}-50=50(\sqrt{3}-1) \mathrm{m}\) or $\displaystyle 36.5$ m
(b)
Speed of the ship \(\displaystyle =\frac{50 \text { metres }}{10 \text { minutes }}\)
\(\displaystyle =0.3 \mathrm{~km} / \mathrm{h}\)
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CBSE Class 10 Mathematics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.