Let AB = 120 meters be the width of the road, with two pillars of equal height standing directly opposite each other at points A and B. From point C, which lies on the line AB joining the bases, the tops of the pillars are at points P and Q. The angles of elevation from point C to the tops of the pillars are ∠ACP = 60° and ∠BCQ = 30°, respectively. We need to find the length of AC. From triangle APC: \[ \frac{AP}{AC} = \tan 60^\circ \Rightarrow \frac{AP}{AC} = \sqrt{3} \Rightarrow AC = \frac{AP}{\sqrt{3}} \quad \text{——— (i)} \] From triangle BCQ: \[ \frac{BQ}{BC} = \tan 30^\circ \Rightarrow \frac{BQ}{BC} = \frac{1}{\sqrt{3}} \Rightarrow BC = \sqrt{3} \cdot BQ = \sqrt{3} \cdot AP \quad \text{——— (ii)} \quad [\because AP = BQ] \] We know AB = 120 meters So, \( AC + BC = 120 \) Substituting values from equations (i) and (ii): \[ \frac{AP}{\sqrt{3}} + \sqrt{3} \cdot AP = 120 \Rightarrow AP \left( \sqrt{3} + \frac{1}{\sqrt{3}} \right) = 120 \Rightarrow AP \cdot \frac{3 + 1}{\sqrt{3}} = 120 \Rightarrow AP = \frac{120 \cdot \sqrt{3}}{4} = 30\sqrt{3} \] Now, using equation (i): \[ AC = \frac{30\sqrt{3}}{\sqrt{3}} = 30 \]