1. **Problem Statement:** Points A and B are 60 m apart on level ground with a vertical tower between them. From A, the angle of elevation to the tower top is 30°. From B, the angle of elevation is 45°. Find: a) Horizontal distance from A to the tower base. b) Height of the tower. 2. **Setup and notation:** Let $x$ be the horizontal distance from A to the tower base. Then the distance from B to the tower base is $60 - x$. Let $h$ be the height of the tower. 3. **Using tangent of angles:** From A: $\tan 30^\circ = \frac{h}{x}$ From B: $\tan 45^\circ = \frac{h}{60 - x}$ Recall: $\tan 30^\circ = \frac{1}{\sqrt{3}}$ $\tan 45^\circ = 1$ 4. **Write equations:** From A: $$h = x \times \frac{1}{\sqrt{3}} = \frac{x}{\sqrt{3}}$$ From B: $$h = 1 \times (60 - x) = 60 - x$$ 5. **Equate the two expressions for $h$:** $$\frac{x}{\sqrt{3}} = 60 - x$$ 6. **Solve for $x$:** Multiply both sides by $\sqrt{3}$: $$x = (60 - x) \sqrt{3}$$ $$x = 60\sqrt{3} - x\sqrt{3}$$ Bring terms with $x$ to one side: $$x + x\sqrt{3} = 60\sqrt{3}$$ Factor $x$: $$x(1 + \sqrt{3}) = 60\sqrt{3}$$ $$x = \frac{60\sqrt{3}}{1 + \sqrt{3}}$$ 7. **Rationalize denominator:** $$x = \frac{60\sqrt{3}}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{60\sqrt{3}(1 - \sqrt{3})}{1 - 3} = \frac{60\sqrt{3} - 60 \times 3}{-2} = \frac{60\sqrt{3} - 180}{-2}$$ Divide numerator and denominator: $$x = \frac{60\sqrt{3} - 180}{-2} = -30\sqrt{3} + 90 = 90 - 30\sqrt{3}$$ 8. **Calculate numerical value:** $\sqrt{3} \approx 1.732$ $$x \approx 90 - 30 \times 1.732 = 90 - 51.96 = 38.04 \text{ m}$$ 9. **Find height $h$:** Using $h = \frac{x}{\sqrt{3}}$: $$h = \frac{38.04}{1.732} \approx 21.95 \text{ m}$$ **Final answers:** a) Horizontal distance from A to tower base is approximately $38.04$ m. b) Height of the tower is approximately $21.95$ m.