Q81 / 125 · BITSAT 2009 · Mathematics
A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is \(60^{\circ}\). when he retreats 20 feet from the bank, he finds the angle to be \(30^{\circ}\). The breadth of the river in feet is :
- A15
- B\(15 \sqrt{3}\)
- C\(10 \sqrt{3}\)
- Correct answerD10
Solution
et \(h\) be the height of tree PQ and breadth of river PS be \(x \mathrm{~ft}\). Angle of elevation subtended by a tree is \(60^{\circ}\). Also, when he retreats 20 feet, the angle becomes \(30^{\circ}\).

Also, in \(\triangle \mathrm{PQS}\), \(\begin{aligned} & \tan 60^{\circ}=\frac{h}{x} \\ & \Rightarrow h=\sqrt{3} x \end{aligned}\) and in \(\triangle \mathrm{PQR}\), \(\begin{aligned} & \tan 30^{\circ}=\frac{h}{x+20} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+20} \\ & \Rightarrow x+20=\sqrt{3} h \\ & \Rightarrow x+20=3 x \Rightarrow 2 x=20 \Rightarrow x=10 \end{aligned}\) Hence breadth of river is \(10 \mathrm{ft}\).

Also, in \(\triangle \mathrm{PQS}\), \(\begin{aligned} & \tan 60^{\circ}=\frac{h}{x} \\ & \Rightarrow h=\sqrt{3} x \end{aligned}\) and in \(\triangle \mathrm{PQR}\), \(\begin{aligned} & \tan 30^{\circ}=\frac{h}{x+20} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+20} \\ & \Rightarrow x+20=\sqrt{3} h \\ & \Rightarrow x+20=3 x \Rightarrow 2 x=20 \Rightarrow x=10 \end{aligned}\) Hence breadth of river is \(10 \mathrm{ft}\).
Hence, the correct answer is option: 'D'.