a man standing on the bank of a river observes that the angle of elevation of a tree on the - Maths - Some Applications of Trigonometry

Question

a man standing on the bank of a river observes that the angle of
elevation of a tree on the opposite bank is 60 degree when he moves
50m away from the bank , he finds the angle of elevation to be 30
degree
Find the (1)the width of the river (2)the height of the tree

Santosh Kumar · Accepted Answer

@JhalakJauhari: Good effort! Keep posting! The information given in the problem can be represented as,

Let AB be the tree of height h m Also, let the width of the river, BC be x m Suppose C and D be the initial and final position of the man In Î”ABD, $\tan 30 ° = \frac{AB}{BD} = \frac{h}{BC + CD} ∴ \frac{1}{\sqrt{3}} = \frac{h}{x + 50} \Rightarrow h = \frac{x + 50}{\sqrt{3}} (1)$ In Î”ABC, $\tan 60 ° = \frac{AB}{BC} ∴ \sqrt{3} = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{x + 50}{\sqrt{3} x} [Using (1)] \Rightarrow \sqrt{3} \times \sqrt{3} x = x + 50 \Rightarrow 3 x = x + 50 \Rightarrow 2 x = 50 \Rightarrow x = 25 m$ âˆ´ Width of the river = 25 m From (1), we have $h = \frac{x + 50}{\sqrt{3}} = \frac{25 + 50}{\sqrt{3}} = \frac{75}{\sqrt{3}} = \frac{75}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{75 \sqrt{3}}{3} = 25 \sqrt{3} m$ âˆ´ Height of the tree = $25 \sqrt{3}$ m

a man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60 degree. when he moves 50m away from the bank , he finds the angle of elevation to be 30 degree. Find the (1)the width of the river (2)the height of the tree.

a man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is 60 degree. when he moves 50m away from the bank , he finds the angle of elevation to be 30 degree.

Find the (1)the width of the river (2)the height of the tree.