Question

answer ittt


ch.- height and distance

class 10 ​

Answer 1

Considering the formation of right triangle with breaking of trees' branch & touching the ground.

Here,

• Base = 15 m
• Angle between base & hypotenuse = 45°

Applying trigonometric ratios, we get,

$\text{tan 45°= } \bf{ \frac{perpendicular}{base} } \\ \\ = > \rm 1 = \frac{perpendicular}{15} \\ \\ \rm = > perpendicular = 15 \times 1 = \green{15 \:}{ m}$

$\text{sec 45° =} \bf \frac{hypotenuse}{base} \\ \\ \rm = > \sqrt{2} = \frac{hypotenuse}{15} \\ \rm = > hypotenuse = 15 \times \sqrt{2} = \green{15 \sqrt{2} } \: m$

And,

• Total height of tree = broken part(hypotenuse) + unbroken part(perpendicular)

=> 15 + 15√2

=> 15(1 + √2) m

=> 36.21 m (approx.)