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Case Study - 2
37. 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, $20$ 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 $10$ m/s and $5$ 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 ?
37. 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, $20$ 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 $10$ m/s and $5$ 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 ?
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(i) $\frac{20}{AD} = \tan 30^\circ \Rightarrow AD = 20\sqrt{3} = 34.6$ m ($\frac{1}{2} + \frac{1}{2}$ marks)
(ii) $\frac{20}{DB} = \tan 45^\circ \Rightarrow DB = 20$ m. Width $= 34.6 + 20 = 54.6$ m ($\frac{1}{2} + \frac{1}{2}$ marks)
(iii) (a) Time $A = \frac{34.6}{10} = 3.46$ s. Time $B = \frac{20}{5} = 4$ s. Boat A reaches earlier by $0.54$ s. (1 + $\frac{1}{2} + \frac{1}{2}$ marks) OR (b) Dist $A = 30$ m, Dist $B = 15$ m. Dist between them $= 54.6 - (30+15) = 9.6$ m ($\frac{1}{2} + \frac{1}{2} + 1$ marks)
(ii) $\frac{20}{DB} = \tan 45^\circ \Rightarrow DB = 20$ m. Width $= 34.6 + 20 = 54.6$ m ($\frac{1}{2} + \frac{1}{2}$ marks)
(iii) (a) Time $A = \frac{34.6}{10} = 3.46$ s. Time $B = \frac{20}{5} = 4$ s. Boat A reaches earlier by $0.54$ s. (1 + $\frac{1}{2} + \frac{1}{2}$ marks) OR (b) Dist $A = 30$ m, Dist $B = 15$ m. Dist between them $= 54.6 - (30+15) = 9.6$ m ($\frac{1}{2} + \frac{1}{2} + 1$ marks)