Question

how to form a figure????? A man sitting at a height of 20m on a tall tree on a small tree on a small island in the middle of a river observes two poles directly opposite to each other on the two banks of the river and in line with the foot of the tree. If the angle of depressionof the feet of the poles from a point at which the man is sitting on the tree on either side of river is 60* and 30* respectively. Find the width of the river.

Answer 1

$Height \: of \:the \:Tree (AC) = 20 \:m$

$B \:and \: C \: are \: poles \: directly \\ opposite\: to \: each \:other \:on \:two \:banks \\of \:the \:river .$

$Let \: BC = x \: m \: and \: CD = y \:m$

$Distance \: between \: foot\: of \\two \: poles (BD)= ( x + y ) m$

$In\: \triangle ABC, \\</p><p>tan 60\degree = \frac{AC}{BC} \\\implies \sqrt{3} = \frac{20}{x} \\\implies x = \frac{20}{\sqrt{3}}\: --(1)$

$In\: \triangle ACD, \\</p><p>tan 30\degree = \frac{AC}{CD} \\\implies \frac{1}{\sqrt{3}} = \frac{20}{y} \\\implies y = 20\sqrt{3}\: --(2)$

$Distance \: between \: foot\: of \\two \: poles (BD)= ( x + y ) m\\= \frac{20}{\sqrt{3}} + 20\sqrt{3}\\= \frac{20+20\sqrt{3} \times \sqrt{3}}{\sqrt{3}}\\= \frac{20 + 20\times 3}{\sqrt{3}}\\= \frac{20+60}{\sqrt{3}}\\= \frac{80}{\sqrt{3}}\\= \frac{80\sqrt{3}}{\sqrt{3}\times \sqrt{3}}\\= \frac{80\sqrt{3}}{3}\\= \frac{138.56}{3} \\= 46.19 \:m$

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Answer 2

Answer:

80root3/3

Step-by-step explanation:

applications of trigonometry