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 ${60}^o$. When he moves $40m$ away from the bank, he finds the angle of elevation to be ${30}^o$. Calculate :
(i) the width of the river.
(ii) the height of the tree.

Answer 1

$\begin{array}{c}\Delta ABC\\ \tan {60}^{\circ }=\frac{h}{x}\\ \sqrt{3}=\frac{h}{x}\\ h=\sqrt{3}x\\ \Delta DAB\\ \tan {30}^{\circ }=\frac{h}{\left(40+x\right)}\\ \left(40+x\right)=\sqrt{3}h\\ \left(40+x\right)=\sqrt{3}x\sqrt{3}x\\ \left(40+x\right)=3x\\ 2x=40\\ x=20\end{array}$

width of the river is $20m$

$h=\sqrt{3}x=20\sqrt{3}$

Height of the tree is $20\sqrt{3}$