Question

A man observed the top of a tower at a distance a from its base at an angle of elevation 60°. He saw the top of the tower at an angle of elevation 30° from a point at the distance b from the base. Prove that height of the tower h = √ab

Answer 1

\(\frac{h}{b}=tan60^\circ=\sqrt{3}\)

\(\frac{h}{a}=tan30^\circ=\frac{1}{\sqrt{3}}\)

\(\frac{h}{b}\times\frac{h}{a}=\sqrt{3}\times\frac{1}{\sqrt{3}}=1\)

\(\frac{h^2}{ab}=1,\)

\(h=\sqrt{ab}\)