Let AB represent the height of the tree. Suppose the tree broke at point C and its top touches the ground at D.
AC is the broken part of the tree which takes position CD such that ∠CDB = 60°
∴ AC = CD …(i)
BD = 20m
In right-angled ∆CBD,
tan 60° = BC/BD … [By definition]
∴ √3 = BC/20
∴ BC = 20√3m
Also, cos 60° = BC/CD … [By definition]
∴ 1/2 = 20/CD
∴ CD = 20 × 2 = 40 m
∴ AC = 40 m …[From(i)]
Now, AB = AC + BC ….[A – C – B]
= 40 + 20√3
= 40 + 20 × 1.73
= 40 + 34.6
= 74.6
∴ The height of the tree is 74.6 m.
