Let AB be a 16-meter high building, with A as the rooftop and B as the base. From points A and B, the angles of elevation to the top of a temple at point P are ∠PAM = 45° and ∠PBQ = 60°, respectively. From triangle PQB: \[ \frac{PQ}{QB} = \tan 60^\circ = \sqrt{3} \Rightarrow PQ = \sqrt{3} \cdot QB \] From triangle PMA: \[ \frac{PM}{MA} = \tan 45^\circ = 1 \Rightarrow PM = MA = QB \] Now, \[ PQ - PM = MQ \Rightarrow PQ - PM = AB \quad [\text{since } MQ = AB] \Rightarrow \sqrt{3} \cdot QB - QB = 16 \Rightarrow QB(\sqrt{3} - 1) = 16 \Rightarrow QB = \frac{16}{\sqrt{3} - 1} \] Rationalizing the denominator: \[ QB = \frac{16(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{16(\sqrt{3} + 1)}{3 - 1} = \frac{16(\sqrt{3} + 1)}{2} = 8(\sqrt{3} + 1) \] Therefore, \[ PQ = \sqrt{3} \cdot QB = \sqrt{3} \cdot 8(\sqrt{3} + 1) = 24 + 8\sqrt{3} \] So, the height of the temple is \(24 + 8\sqrt{3}\) meters, and the horizontal distance from the building to the temple is \(8(\sqrt{3} + 1)\) meters.