Question

The upper part of a tree broken by the wind makes an angle of ${30}^{\circ }$ with the ground and the distance from the foot to the point where the top of the tree meets the ground is $8$ metres . What was the height of the tree?

Answer 1

Let $l$ and $h$ be the length of broken and unbroken part of the tree.

$tan{30}^0=\left(\frac{h}{8}\right)orh=8tan{30}^0=\left(\frac{8}{\sqrt{3}}\right)cos{30}^0=\left(\frac{8}{l}\right)orl=\left(\frac{8}{cos{30}^0}\right)=\left(\frac{8}{\left(\frac{\sqrt{3}}{2}\right)}\right)=\left(\frac{16}{\sqrt{2}}\right)\therefore totalheightoftree=\left(\frac{8}{\sqrt{3}}\right)+\left(\frac{16}{\sqrt{3}}\right)=\left(\frac{24}{\sqrt{3}}\right)=8\sqrt{3}$ meters