1. **Problem Statement:** Two points are 40 m apart along a straight path observing a flagpole. The angle of elevation from the first point is 30° and from the second point is 45°. We need to find: a) The height of the flagpole. b) The distance from the first point to the flagpole. 2. **Setup and Formula:** Let the distance from the first point to the base of the flagpole be $x$ meters. The distance from the second point to the flagpole is then $x - 40$ meters (since the points are 40 m apart). Let the height of the flagpole be $h$ meters. Using the tangent of the angles (since tangent relates opposite side to adjacent side in a right triangle): $$\tan(30^\circ) = \frac{h}{x}$$ $$\tan(45^\circ) = \frac{h}{x - 40}$$ 3. **Calculate Tangent Values:** $$\tan(30^\circ) = \frac{1}{\sqrt{3}}$$ $$\tan(45^\circ) = 1$$ 4. **Express $h$ from both equations:** From first point: $$h = x \times \frac{1}{\sqrt{3}} = \frac{x}{\sqrt{3}}$$ From second point: $$h = 1 \times (x - 40) = x - 40$$ 5. **Set the two expressions for $h$ equal:** $$\frac{x}{\sqrt{3}} = x - 40$$ 6. **Solve for $x$:** Multiply both sides by $\sqrt{3}$: $$x = \sqrt{3}(x - 40)$$ $$x = \sqrt{3}x - 40\sqrt{3}$$ Bring terms involving $x$ to one side: $$x - \sqrt{3}x = -40\sqrt{3}$$ $$x(1 - \sqrt{3}) = -40\sqrt{3}$$ $$x = \frac{-40\sqrt{3}}{1 - \sqrt{3}}$$ Multiply numerator and denominator by the conjugate $(1 + \sqrt{3})$ to rationalize: $$x = \frac{-40\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{-40\sqrt{3} - 40 \times 3}{1 - 3} = \frac{-40\sqrt{3} - 120}{-2}$$ $$x = \frac{40\sqrt{3} + 120}{2} = 20\sqrt{3} + 60$$ 7. **Calculate numerical value of $x$:** $$20\sqrt{3} \approx 20 \times 1.732 = 34.64$$ $$x \approx 34.64 + 60 = 94.64 \text{ meters}$$ 8. **Calculate height $h$ using $h = x - 40$:** $$h = 94.64 - 40 = 54.64 \text{ meters}$$ **Final answers:** - Height of the flagpole $h \approx 54.64$ meters. - Distance from the first point to the flagpole $x \approx 94.64$ meters.