Question

A tree is broken by the wind. The top struck the ground at an angle of 30^o and at a distance of 30 m from the root. Find the whole height of the tree.$\left(\sqrt{3}=1.73\right)$

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{Distance}\mathrm{between}\mathrm{the}\mathrm{top}\mathrm{of}\mathrm{the}\mathrm{broken}\mathrm{tree}\mathrm{and}\mathrm{the}\mathrm{root},AC=30\mathrm{m}\mathrm{and}\mathrm{angle}\mathrm{of}\mathrm{elevation}=30° \\ \mathrm{In}△CAB,\tan \theta =\frac{\mathrm{perpendicular}}{\mathrm{base}}=\frac{AB}{AC} \\ \Rightarrow \tan 30°=\frac{AB}{30} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{30} \\ \Rightarrow AB=\frac{30}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=10\times 1.73=17.3\mathrm{m} \\ \mathrm{Also},\cos \theta =\frac{\mathrm{base}}{\mathrm{hypotenuse}}=\frac{AC}{BC} \\ \Rightarrow \cos 30°=\frac{30}{BC} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{30}{BC} \\ \Rightarrow BC=\frac{60\sqrt{3}}{3}=20\sqrt{3}\mathrm{m} \\ \Rightarrow BC=20\times 1.73=34.6\mathrm{m} \\ \therefore \mathrm{Original}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{tree}=AB+BC=(17.3+34.6)\mathrm{m}=51.9\mathrm{m} \end{gathered}$$