We are given that sector $OPAQ$ with radius $r$ cm is folded into a cone with a base radius of 3 cm and a height of 6 cm. The volume of the cone is 50 cm³, and we are tasked with finding the value of $\theta$, the angle of the sector.

Step 1: Recall the formula for the volume of a cone

The volume $V$ of a cone is given by the formula: $V = \frac{1}{3} \pi r_{\text{base}}^2 h$ where $r_{\text{base}}$ is the radius of the base and $h$ is the height of the cone. Plugging in the given values: $50 = \frac{1}{3} \pi (3^2)(6)$ $50 = \frac{1}{3} \pi (9)(6) = 18\pi$ This implies that: $50 = 18\pi$ Now solve for $\pi \approx 3.14$: $50 = 18 \times 3.14 = 56.52$ There seems to be a mismatch, suggesting we should look to the