Question

a vertically straight tree 15m high is broken by the wind in such a way that it's top just touches the ground and maks an angle of 60° with the ground.at what height from the ground did the tree break?​

Answer 1

Answer:

$30\sqrt{3} - 45$

or

6.96152422707 approx

Step-by-step explanation:

the tree forms a right angled triangle.

sinθ = opposite / hypotenuse

sin 60 = $\frac{\sqrt{3} }{2\\}$

so now we know total height is 15m that is hypotenuse+height

we know ratio between height and hypotenuse is $\sqrt{3} : 2$

so $\sqrt{3} x$ + 2x = 15

x = 15$\frac{15}{2+\sqrt{3} }$

if you faactorise the equation you get

x = 30 - 15$\sqrt{3}$

so height from ground where it broke = $\sqrt{3}$(30-15$\sqrt{3}$

= 30$30\sqrt{3} - 45$