1. **Problem Statement:** A lighthouse 50 m tall observes a ship moving from point P to Q. The angles of depression from the top of the lighthouse to the ship are 30° at P and 45° at Q. We need to find: (i) Distance of the ship from the base of the lighthouse at Q. (ii) Measures of angles \(\angle PBA\) and \(\angle QBA\). (iii) (a) Distance traveled by the ship from P to Q. (iii) (b) Speed of the ship from Q to A if it takes 10 minutes. 2. **Key Concepts and Formulas:** - The angle of depression equals the angle of elevation from the ship to the lighthouse base. - Using right triangle trigonometry, for height \(h=50\) m and horizontal distance \(d\), \(\tan(\theta) = \frac{h}{d}\). - Speed formula: \(\text{speed} = \frac{\text{distance}}{\text{time}}\). 3. **Step (i): Distance at Q** - Given angle of depression at Q is 45°. - Using \(\tan 45^\circ = 1 = \frac{50}{d_Q} \Rightarrow d_Q = 50\) m. 4. **Step (ii): Angles \(\angle PBA\) and \(\angle QBA\)** - By alternate interior angles, \(\angle PBA = 30^\circ\) and \(\angle QBA = 45^\circ\) (given angles of depression). 5. **Step (iii)(a): Distance traveled from P to Q** - At P, angle of depression is 30°. - \(\tan 30^\circ = \frac{50}{d_P} = \frac{1}{\sqrt{3}} \Rightarrow d_P = 50\sqrt{3} \approx 86.60\) m. - Distance traveled = \(d_P - d_Q = 86.60 - 50 = 36.60\) m. 6. **Step (iii)(b): Speed from Q to A** - Distance from Q to A is \(d_Q = 50\) m. - Time = 10 minutes = \(\frac{10}{60} = \frac{1}{6}\) hours. - Speed = \(\frac{50}{1000} \div \frac{1}{6} = 0.05 \times 6 = 0.3\) km/h. **Final answers:** (i) Distance at Q = 50 m. (ii) \(\angle PBA = 30^\circ\), \(\angle QBA = 45^\circ\). (iii)(a) Distance traveled = 36.60 m. (iii)(b) Speed = 0.3 km/h.