A pole got between by the wind whose top meets the ground at angle of 60° at a distance of 190 meters from the foot.  What was the total height of the pole? (\(\sqrt {3}\) = 1.732)

709.08

Tan 60° = \(\frac{AB}{190}\)

AB = 190\(\sqrt {3}\) .... (i)

Sin 60° = \(\frac{AB}{AC}\)

\(\frac{\sqrt {3}}{2}\) = \(\frac{190\sqrt {3}}{AC}\)

AC = 380 m ... (ii)

height of the pole = AB + AC = 190\(\sqrt {3}\) + 380 = 190 (\(\sqrt {3}\) + 2)

= 190 (1.732 + 2) = 190 × 3.732 = 709.08