1. **Problem statement:** From a point on the ground in front of a tower, the angle of elevation to the top of a 6 m high flagstaff on the tower is 60° and to the bottom of the flagstaff (top of the tower) is 45°. Find the height of the tower and the distance from the point to the base of the tower. 2. **Define variables:** Let the height of the tower be $h$ meters and the distance from the point of observation to the base of the tower be $d$ meters. 3. **Use trigonometric relations:** - The bottom of the flagstaff is at height $h$. - The top of the flagstaff is at height $h + 6$. 4. **From the angle of elevation to the bottom (45°):** $$\tan 45^\circ = \frac{h}{d}$$ Since $\tan 45^\circ = 1$, we get $$1 = \frac{h}{d} \implies h = d$$ 5. **From the angle of elevation to the top (60°):** $$\tan 60^\circ = \frac{h + 6}{d}$$ Since $\tan 60^\circ = \sqrt{3}$, we have $$\sqrt{3} = \frac{h + 6}{d}$$ 6. **Substitute $h = d$ into the above:** $$\sqrt{3} = \frac{d + 6}{d} = 1 + \frac{6}{d}$$ 7. **Solve for $d$:** $$\sqrt{3} - 1 = \frac{6}{d} \implies d = \frac{6}{\sqrt{3} - 1}$$ 8. **Rationalize the denominator:** $$d = \frac{6}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{6(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{6(\sqrt{3} + 1)}{3 - 1} = \frac{6(\sqrt{3} + 1)}{2} = 3(\sqrt{3} + 1)$$ 9. **Calculate $h$:** Since $h = d$, $$h = 3(\sqrt{3} + 1)$$ **Final answers:** - Height of the tower: $$h = 3(\sqrt{3} + 1) \approx 3(1.732 + 1) = 3(2.732) = 8.196 \text{ meters}$$ - Distance from the point to the base of the tower: $$d = 3(\sqrt{3} + 1) \approx 8.196 \text{ meters}$$