Question

The upper part of a tree broken over by the wind makes an angle of ${30}^{\circ }$ with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. Using sine rule, find the height of the tree.

Answer 1

$BC=15m,AB=h\text{From the diagram we can calculate,}\angle A={60}^{\circ }\text{Using sine rule,}\frac{sinA}{15}=\frac{sinC}{h}\Rightarrow \frac{sin60}{15}=\frac{sin30}{h}\Rightarrow \frac{\sqrt{3}}{2\times 15}=\frac{1}{2\times h}\Rightarrow \frac{\sqrt{3}}{15}=\frac{1}{h}\Rightarrow h=\frac{15}{\sqrt{3}}\Rightarrow h=5\sqrt{3}$

$h=5\sqrt{3}Also,\frac{15}{sin{60}^{\circ }}=\frac{y}{sin{90}^{\circ }}\Rightarrow \frac{\frac{15}{\sqrt{3}}}{2}=y\Rightarrow y=\frac{30}{\sqrt{3}}=10\sqrt{3}\text{So,the height of the true is}h+y=5\sqrt{3}+10\sqrt{3}m=15\sqrt{3}m$