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Two motorboats $A$ and $B$ are waiting at the opposite banks of a river in order to reach the opposite side. From a point $P$ on the bridge, $20 \text{ m}$ above the river, the angles of depression of the boats are $30^\circ$ and $45^\circ$ respectively, as shown in the figure given below. Both the boats leave at the same time at the speed of $10 \text{ m/s}$ and $5 \text{ m/s}$, respectively. Based on the above information, answer the following questions :
(i) Find the distance travelled by boat $A$ to reach point $D$ in the river, vertically below the point $P$. (Use $\sqrt{3} = 1.73$)
(ii) What is the width of the river ?
(iii) (a) Which boat will reach point $D$ first, and how much earlier, than the other boat ?
OR
(b) What is the distance between the two boats after $3$ seconds ?
(i) Find the distance travelled by boat $A$ to reach point $D$ in the river, vertically below the point $P$. (Use $\sqrt{3} = 1.73$)
(ii) What is the width of the river ?
(iii) (a) Which boat will reach point $D$ first, and how much earlier, than the other boat ?
OR
(b) What is the distance between the two boats after $3$ seconds ?
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Solution: (i) $\frac{20}{AD} = \tan 30^\circ \Rightarrow AD = 20\sqrt{3} = 34.6 \text{ m}$
(ii) $\frac{20}{DB} = \tan 45^\circ \Rightarrow DB = 20 \text{ m}$
Width of river $= 34.6 + 20 = 54.6 \text{ m}$
(iii) (a) Time taken by boat $A = \frac{34.6}{10} = 3.46 \text{ seconds}$
Time taken by boat $B = \frac{20}{5} = 4 \text{ seconds}$
Boat $A$ will reach earlier by $0.54 \text{ seconds}$
OR
(iii) (b) Distance covered by boat $A$ in $3 \text{ seconds} = 3 \times 10 = 30 \text{ m}$
Distance covered by boat $B$ in $3 \text{ seconds} = 3 \times 5 = 15 \text{ m}$
Distance between them after $3 \text{ seconds} = 54.6 - (30 + 15) = 9.6 \text{ m}$
(ii) $\frac{20}{DB} = \tan 45^\circ \Rightarrow DB = 20 \text{ m}$
Width of river $= 34.6 + 20 = 54.6 \text{ m}$
(iii) (a) Time taken by boat $A = \frac{34.6}{10} = 3.46 \text{ seconds}$
Time taken by boat $B = \frac{20}{5} = 4 \text{ seconds}$
Boat $A$ will reach earlier by $0.54 \text{ seconds}$
OR
(iii) (b) Distance covered by boat $A$ in $3 \text{ seconds} = 3 \times 10 = 30 \text{ m}$
Distance covered by boat $B$ in $3 \text{ seconds} = 3 \times 5 = 15 \text{ m}$
Distance between them after $3 \text{ seconds} = 54.6 - (30 + 15) = 9.6 \text{ m}$