The upper part of a tree broken by wind, falls to the ground without being detached. The top of the broken part touches the ground at an angle of 38° 30' at a point 6 m from the foot of the tree. Calculate :

(i) the height at which the tree is broken.

(ii) the original height of the tree correct to two decimal places.

(i) Let ACB be the tree. When broken at point C by the storm, let its top A touch the ground so that ∠CAB = 38° 30' and AB = 6 m.

From right angled △ABC, we get

$\Rightarrow \text{tan 38° 30'} = \dfrac{BC}{AB} \\[1em] \Rightarrow 0.7954 = \dfrac{BC}{6} \\[1em] \Rightarrow BC = 0.7954 \times 6 \\[1em] \Rightarrow BC = 4.77$

Hence, the tree is broken at a height of 4.77 m

(ii) From right angled △ABC, we get

$\Rightarrow \text{cos 38° 30}' = \dfrac{AB}{AC} \\[1em] \Rightarrow 0.7826 = \dfrac{6}{AC} \\[1em] \Rightarrow AC = \dfrac{6}{0.7826} \\[1em] \Rightarrow AC = 7.67$

∴ The height of the tree = BC + AC = 4.77 + 7.67 = 12.44

Hence, the original height of the tree is 12.44 metres.