Distance between the foot of tree and the point of contact of the top of the tree on the ground = 6 cm.
Let the length of the remaining part be = h m.
Let the length of the broken part be = x m.
Angle made by the broken part with the ground = 30°.
From the figure tan 30° = \(\frac{h}{6}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac{h}{6}\)
∴ h = \(\frac{6}{\sqrt{3}}\) = \(\frac{3\times2}{\sqrt{3}}\) = 2√3 m
Also cos 30° = \(\frac{6}{x}\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac{6}{x}\)
⇒ x = \(\frac{6\times2}{\sqrt{3}}\) = \(\frac{3\times2\times2}{\sqrt{3}}\) = 4√3
∴ Height of the tree = broken part + remaining part
= x + h
= 2√3 + 4√3
= 6√3 m
= 6 × 1.732
≃ 10.392 m.
