The heights of two pillars are \(h_1\) meters and \(h_2\) meters respectively. If the angle of elevation from the base of the second pillar to the top of the first pillar is 60°, and the angle of elevation from the base of the first pillar to the top of the second pillar is 45°, show that \(h_1^2 = 3h_2^2\).

Q.The heights of two pillars are \(h_1\) meters and \(h_2\) meters respectively. If the angle of elevation from the base of the second pillar to the top of the first pillar is 60°, and the angle of elevation from the base of the first pillar to the top of the second pillar is 45°, show that \(h_1^2 = 3h_2^2\).

Let the heights of the two pillars be AB = \(h_1\) meters and PQ = \(h_2\) meters. From the base of the second pillar at point Q, the angle of elevation to the top of the first pillar at point A is \(\angle AQB = 60^\circ\), and from the base of the first pillar at point B, the angle of elevation to the top of the second pillar at point P is \(\angle PBQ = 45^\circ\). From triangle ABQ, \[ \tan 60^\circ = \frac{AB}{BQ} \Rightarrow \sqrt{3} = \frac{h_1}{BQ} \Rightarrow BQ = \frac{h_1}{\sqrt{3}} \quad \text{(i)} \] From triangle PQB, \[ \tan 45^\circ = \frac{PQ}{BQ} \Rightarrow 1 = \frac{h_2}{BQ} \Rightarrow BQ = h_2 \quad \text{(ii)} \] From equations (i) and (ii), \[ \frac{h_1}{\sqrt{3}} = h_2 \Rightarrow h_1 = \sqrt{3} h_2 \Rightarrow h_1^2 = 3 h_2^2 \quad \text{(Proved)} \]