Let \(x\) be the distance between the base of the tower and the point.

The angle subtended by the pole is \(\phi\) and the angle of elevation of the top of the tower is \( \theta\). But these two angles are equal, \(\phi = \theta\).

\(\tan{\theta} = \frac{30}{x}\)

\( \tan{2\theta} = \frac{30+34}{x}\): since \(\theta = \phi\)

By trigometric identities:

\(\tan{2\theta} = \frac{2\tan{\theta}}{1-(\tan{\theta})^2}\)

\(\frac{64}{x} = \frac{2\frac{30}{x}}{1-(\frac{30}{x})^2}\)

\(\frac{64}{x} = \frac{\frac{60}{x}}{1-\frac{900}{x^2}}\)

\(64 = \frac{60x^2}{x^2-900}\)

\(64(x^2-900) = 60x^2\)

\(64x^2-57600 = 60x^2\)

\(4x^2 = 57600\)

\(x =120 ~ft\) Answer

Let \(x\) be the distance between the base of the tower and the point.

The angle subtended by the pole is \(\phi\) and the angle of elevation of the top of the tower is \( \theta\). But these two angles are equal, \(\phi = \theta\).

\(\tan{\theta} = \frac{30}{x}\)

\( \tan{2\theta} = \frac{30+34}{x}\): since \(\theta = \phi\)

By trigometric identities:

\(\tan{2\theta} = \frac{2\tan{\theta}}{1-(\tan{\theta})^2}\)

\(\frac{64}{x} = \frac{2\frac{30}{x}}{1-(\frac{30}{x})^2}\)

\(\frac{64}{x} = \frac{\frac{60}{x}}{1-\frac{900}{x^2}}\)

\(64 = \frac{60x^2}{x^2-900}\)

\(64(x^2-900) = 60x^2\)

\(64x^2-57600 = 60x^2\)

\(4x^2 = 57600\)

\(x =120 ~ft\) Answer