設觀景台頂端為 $T$，則 $\overline{OT} = 150$。 依俯角條件可得 $A, B, C$ 到 $O$ 點的距離： $\overline{OA} = 150 \cot 30^\circ = 150\sqrt{3}$ $\overline{OB} = 150 \cot 60^\circ = \dfrac{150}{\sqrt{3}} = 50\sqrt{3}$ $\overline{OC} = 150 \cot 45^\circ = 150$ 在 $\triangle OAB$ 中，$C$ 為 $\overline{AB}$ 之中點，由中線定理知： $\overline{OA}^2 + \overline{OB}^2 = 2(\overline{OC}^2 + \overline{AC}^2)$ $(150\sqrt{3})^2 + (50\sqrt{3})^2 = 2(150^2 + \overline{AC}^2)$ $67500 + 7500 = 2(22500 + \overline{AC}^2)$ $75000 = 45000 + 2\overline{AC}^2 \implies 2\overline{AC}^2 = 30000 \implies \overline{AC}^2 = 15000$ 在 $\triangle OAB$ 中，由餘弦定理求 $\cos\angle AOB$： 設 $\angle AOB = \theta$，且 $\overline{AB} = 2\overline{AC} \implies \overline{AB}^2 = 4 \times 15000 = 60000$ $60000 = (150\sqrt{3})^2 + (50\sqrt{3})^2 - 2(150\sqrt{3})(50\sqrt{3})\cos\theta$ $60000 = 67500 + 7500 - 45000\cos\theta$ $45000\cos\theta = 15000 \implies \cos\theta = \dfrac{1}{3}$ 則 $\sin\theta = \sqrt{1 - (1/3)^2} = \dfrac{2\sqrt{2}}{3}$。 三角形面積 $Area = \dfrac{1}{2} \overline{OA} \cdot \overline{OB} \cdot \sin\theta$ $= \dfrac{1}{2} (150\sqrt{3}) (50\sqrt{3}) \left(\dfrac{2\sqrt{2}}{3}\right) = \dfrac{1}{2} \times 22500 \times \dfrac{2\sqrt{2}}{3} = 7500\sqrt{2}$。