Question

A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is ${60}^{\circ }$. When he moves 30 metres away from the bank, he finds the angle of elevation to be ${30}^{\circ }$. Find the height of the tree and the width of the river. [Take $\sqrt{3}=1.732$] [4 MARKS]

Answer 1

Let AB be the tree and AC be the river.

Let C and D be the two positions of the person.

Then, $\angle ACB={60}^{\circ },\angle ADB={30}^{\circ },\angle DAB={90}^{\circ }$ and CD = 30 m

Let AB = h metres and AC = x metres.

From right $\Delta CAB$, we have

$\frac{AC}{AB}=cot{60}^{\circ }=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{x}{h}=\frac{1}{\sqrt{3}}\Rightarrow x=\frac{h}{\sqrt{3}}$ ....... (i)



From right $\Delta DAB$, we have

$\frac{AD}{AB}=cot{30}^{\circ }=\sqrt{3}$

$\Rightarrow \frac{x+30}{h}=\sqrt{3}\Rightarrow x=\sqrt{3}h-30$ ....... (ii)

Equating the values of x from (i) and (ii), we get

$\frac{h}{\sqrt{3}}=\sqrt{3}h-30\Rightarrow h=3h-30\sqrt{3}$

$\Rightarrow 2h=30\sqrt{3}\Rightarrow h=15\sqrt{3}=15\times 1.732=25.98$

Putting $h=15\sqrt{3}$ in (i), we get $x=\frac{15\sqrt{3}}{\sqrt{3}}=15$.

[$\frac{1}{2}$ MARK]

Let h = height of tree

x = width of river

$\begin{array}{c}In\Delta BAC\left[1MARK\right]\\ tan60=\frac{h}{x}\\ \sqrt{3}=\frac{h}{x}\dots \dots \left(1\right)\left[\frac{1}{2}MARK\right]\\ \\ \\ \\ \\ \\ \\ \\ \end{array}|\begin{array}{c}In\Delta DAC\left[1MARK\right]\\ tan30=\frac{h}{x+30}\\ \frac{1}{\sqrt{3}}=\frac{h}{\frac{h}{\sqrt{3}}+30}\left(from\right(1\left)\right)\\ \frac{1}{\sqrt{3}}=\frac{h\sqrt{3}}{h+30\sqrt{3}}\\ h+30\sqrt{3}=3h\\ 2h=30\sqrt{3}\\ \therefore x=\frac{15\sqrt{3}}{3}=15m\\ h=15\sqrt{3}m\\ h=15\left(1.732\right)\\ =25.98m\left[\frac{1}{2}MARK\right]\end{array}$

Hence, the height of the tree is 25.98 m and the width of the river is 15 metres.