1. **Problem statement:** We have two observation points 40 m apart on a straight line. From the first point, the angle of elevation to the top of the flagpole is 30° and from the second point, it is 45°. We need to find: a) The height of the flagpole. b) The distance from the first point to the base of the flagpole. 2. **Setup and notation:** Let the distance from the first point to the base of the flagpole be $x$ meters. Then the distance from the second point to the base is $x - 40$ meters (since the points are 40 m apart). Let the height of the flagpole be $h$ meters. 3. **Using trigonometry:** The tangent of the angle of elevation relates height and distance: $$\tan(\theta) = \frac{h}{\text{distance}}$$ From the first point (angle 30°): $$\tan(30^\circ) = \frac{h}{x}$$ From the second point (angle 45°): $$\tan(45^\circ) = \frac{h}{x - 40}$$ 4. **Values of tangent:** $$\tan(30^\circ) = \frac{1}{\sqrt{3}}$$ $$\tan(45^\circ) = 1$$ 5. **Write equations:** From first point: $$h = x \times \frac{1}{\sqrt{3}} = \frac{x}{\sqrt{3}}$$ From second point: $$h = 1 \times (x - 40) = x - 40$$ 6. **Equate the two expressions for $h$:** $$\frac{x}{\sqrt{3}} = x - 40$$ 7. **Solve for $x$:** Multiply both sides by $\sqrt{3}$: $$x = \sqrt{3}(x - 40) = \sqrt{3}x - 40\sqrt{3}$$ Rearrange: $$x - \sqrt{3}x = -40\sqrt{3}$$ $$x(1 - \sqrt{3}) = -40\sqrt{3}$$ $$x = \frac{-40\sqrt{3}}{1 - \sqrt{3}}$$ Multiply numerator and denominator by the conjugate $(1 + \sqrt{3})$: $$x = \frac{-40\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{-40\sqrt{3} - 40 \times 3}{1 - 3} = \frac{-40\sqrt{3} - 120}{-2}$$ Simplify denominator: $$x = \frac{-40\sqrt{3} - 120}{-2} = 20\sqrt{3} + 60$$ 8. **Calculate numerical value:** $$20\sqrt{3} \approx 20 \times 1.732 = 34.64$$ So, $$x \approx 34.64 + 60 = 94.64 \text{ meters}$$ 9. **Find height $h$:** Using $h = \frac{x}{\sqrt{3}}$: $$h = \frac{94.64}{1.732} \approx 54.6 \text{ meters}$$ 10. **Final answers:** a) Height of the flagpole $h \approx 54.6$ meters. b) Distance from the first point to the base $x \approx 94.6$ meters.