1. **Problem statement:** We have a right cone monument with vertical height $20$ m. Oliver stands $5$ m from the base, his eye level is $1.8$ m above ground, and the angle of elevation to the cone's vertex is $58^\circ$. We need to find: (a) the radius of the base of the cone. (b) the volume of the cone. 2. **Step (a) - Find the radius:** - The vertical height from Oliver's eye level to the vertex is $20 - 1.8 = 18.2$ m. - The horizontal distance from Oliver to the cone's base is $5$ m. - Using the angle of elevation $58^\circ$, the horizontal distance from the cone's center to the base edge (radius $r$) can be found by first finding the horizontal distance from Oliver to the vertex projection on the ground. 3. **Calculate horizontal distance from Oliver to vertex projection:** Using tangent: $$\tan(58^\circ) = \frac{18.2}{d} \implies d = \frac{18.2}{\tan(58^\circ)}$$ Calculate $d$: $$d = \frac{18.2}{\tan(58^\circ)} \approx \frac{18.2}{1.6003} \approx 11.37 \text{ m}$$ 4. **Find radius $r$:** Oliver is $5$ m from the base, so radius is: $$r = d - 5 = 11.37 - 5 = 6.37 \text{ m}$$ 5. **Step (b) - Find volume of the cone:** Volume formula for a cone: $$V = \frac{1}{3} \pi r^2 h$$ Substitute $r = 6.37$ m and $h = 20$ m: $$V = \frac{1}{3} \pi (6.37)^2 (20) = \frac{1}{3} \pi (40.58)(20) = \frac{1}{3} \pi (811.6) = 270.53 \pi \approx 849.5 \text{ m}^3$$ **Final answers:** - Radius of base $r \approx 6.37$ m - Volume $V \approx 849.5$ cubic meters