One sees the top of a tree on a bank of a river at an elevation of $70^{\circ}$ from the other bank. Stepping 20 m back, he sees the top of the tree at an elevation of $55^{\circ}$ . If height of the person is $1.4 m$ then find the width of river. $[t a n 70^{\circ} = 2.75, t a n 55^{\circ} = 1.43]$

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Solution

Let the width of river be AB = X and the person was standing initially at point C and step back 20 m to reach at D. Height CF of person being 1.4 m.

In Right -angled $△$ ABC

$t a n 70^{\circ} = \frac{A B}{A C} = \frac{A B}{x}$

$A B = x . t a n 70^{\circ} = x \times 2.75 = 2.75 x . . . (1)$

In Right -angled $△$ ABD

$t a n 55^{\circ} = \frac{A B}{A D} = \frac{A B}{(x + 20)}$

$A B = (x + 20) . T a n 55^{\circ} = 1.43 (x + 20) . . . (2)$

from 1 and 2 we have

$2.75 x = 1.43 (x + 20) 2.75 x = 1.43 x + 28.6 1.32 x = 28.6 x = 21.66 m Hence, the width of the river is 21.66 m$

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One sees the top of a tree on a bank of a river at an elevation of 70∘ from the other bank. Stepping 20 m back, he sees the top of the tree at an elevation of 55∘. If height of the person is 1.4 m then find the width of river. [tan 70∘=2.75, tan 55∘=1.43]