Question

A boy, 1.4 metres tall standing at the edge of a river bank sees the top of a tree on the edge of the other bank at an elevation of 55${}^o$. Standing back by 3 metres, he sees it at an elevation of 45${}^o$.
(a) Draw a rough figure showing these facts.
(b) How wide is the river and how tall is the tree?
$[\sin {55}^o=0.8192,\cos {55}^o=0.5736,\tan {55}^o=1.4281]$

Answer 1

The rough sketch is as follows
BF represents the tree, DG is the first position of the boy and AE is the new position.
CD represents the river.
We have : AE = DG
Let DC = x, then EC = (x + 3) m.
$tan{45}^o=\frac{CF}{EC}$
$\Rightarrow 1=\frac{CF}{x+3}$
$\Rightarrow CF=x+3$ ...... (i)
$tan{55}^o=\frac{CF}{DC}$
$\Rightarrow 1.4281=\frac{CF}{x}$
$\Rightarrow CF=1.4281\times x$ ..... (ii)
From (i) and (ii), we have
$1.4281x=x+3$
$0.4281x=3\Rightarrow x=\frac{3}{0.4281}\sim 7m$
Width of the river = CD = 7 m
Height of the tree = BF = BC = CF = (1.4 + 7 + 3)m = 11.4 m