Finding the length of the point from the ground where the tree was broken A vertically straight tree, $15$ $m$ high, is broken by the wind in such a way that its top just touches the ground and makes an angle of $60^{\circ}$ with the ground. At what height from the ground did the tree break? Attached below is the diagram that I have drawn. Kindly check that along with what I've tried already. $ $m$ In $\triangle ACB$, we have:- $tan$ ${60}^{\circ}$=$\frac{AC}{AB}$ $\sqrt3$=$\frac{15-x}{AB}$ But, I cant carry on further.

Let $n$ be the length of the stick which is broken off. The length of the stick which is not broken off will therefore be $15-n$. Note that the stick which is broken off is the hypotenuse of a right triangle, while the stick which is not broken off is one of the legs. Also, notice that the sine function is the **opposite over the hypotenuse**. So if $n$ is the hypotenuse and $15-n$ is the side opposite the angle, we get that

$$\frac{15-n}{n} = \sin{60^o}$$

which, using the unit circle evaluates to $\sqrt{3/4}$. Now we get the equation

$$\frac{15-n}{n} = \sqrt{3/4}$$

which simplifies to

$$15-n = n\sqrt{3/4}$$ $$15 = n + n\sqrt{3/4}$$ $$15 = n(1 + \sqrt{3/4})$$ $$n=15(1+\sqrt{3/4})$$

Therefore, the hypotenuse is $15(1+\sqrt{3/4})$. Solve from here.