Two poles of height 100 m and 100(√3 + 1) m stand upright on a level ground 100 m apart. Find the distance between them.

200 m

AB + RQ = 100 m

Therefore PR = PQ - RQ

                   = 100 (\(\sqrt {3}\) + 1) - 100 = 100 \(\sqrt {3}\) m

tan Θ = \(\frac{PR}{AR}\) = \(\frac{100 \sqrt {3}}{100}\) = \(\sqrt {3}\)

Θ = 60°

sin Θ = \(\frac{PR}{AP}\) ⇒ sin 60° = \(\frac{100 \sqrt {3}}{AP}\)

 ⇒ AP = \(\frac{100 \sqrt {3}}{sin 60°}\) = \(\frac{100 \sqrt {3}}{\sqrt {3}}\) × 2 = 200 m