OPAQ is a sector of a circle with centre 0 and radius cm. The angle at the centre is 0. The sector is then formed into a cone by joining the two radii, OP and OQ, together. The radius of the base of the cone is

r/3$$  $$cm. The cone has a height of 6 cm and a volume of 50π cm'. Find the value of 0.

We are given that OPAQ is a sector of a circle with radius $ r $ cm, and the sector is formed into a cone by joining the two radii. The radius of the base of the cone is $ \frac{r}{3} $ cm, the height of the cone is 6 cm, and the volume of the cone is $ 50\pi $ cm$^3$.

We are asked to find the angle $ \theta $ at the center of the sector.

### Step 1: Volume of the Cone
The formula for the volume $ V $ of a cone is:

\[
V = \frac{1}{3} \pi r_b^2 h
\]

Where:
- $ r_b $ is the radius of the base of the cone,
- $ h $ is the height of the cone.

Substitute the given values:
- $ V = 50\pi $ cm$^3$,
- $ r_b = \frac{r}{3} $,
- $ h = 6 $ cm.

\[
50\pi = \frac{1}{3} \pi \left( \frac{r}{3} \right)^2 (6)
\]

Simplify this equation:
\[
50\pi = \frac{1}{3} \pi \cdot \frac{r^2}{9} \cdot 6
\]

\[
50\pi = \frac{2\pi r^2}{9}
\]

Cancel $ \pi $ from both sides:
\[
50 = \frac{2r^2}{9}
\]

Multiply both sides by 9:
\[
450 = 2r^2
\]

Divide by 2:
\[
r^2 = 225
\]

So:
\[
r = 15 \, \text{cm}
\]

### Step 2: Relation Between the Sector and the Cone
The arc length of the sector forms the circumference of the base of the cone when the sector is rolled into a cone. The formula for the arc length $ L $ of a sector is:

\[
L = r \theta
\]

And the circumference $ C $ of the base of the cone is:

\[
C = 2\pi r_b
\]

Substitute $ r_b = \frac{r}{3} $:
\[
C = 2\pi \times \frac{r}{3} = \frac{2\pi r}{3}
\]

Since the arc length $ L $ of the sector equals the circumference of the base of the cone, we have:

\[
r \theta = \frac{2\pi r}{3}
\]

Cancel $ r $ from both sides:
\[
\theta = \frac{2\pi}{3}
\]

Thus, the angle at the center of the sector is:

\[
\boxed{\frac{2\pi}{3}} \, \text{radians}.
\]

### Final Answer:
The angle $ \theta $ is $ \frac{2\pi}{3} $ radians.

---

Would you like more details or have any questions?

Here are 5 related questions for practice:
1. What is the surface area of the cone in this problem?
2. How would the volume change if the radius of the cone's base doubled?
3. How does the height of the cone affect its volume?
4. What would happen to the angle $ \theta $ if the radius $ r_b $ of the base was halved?
5. How can we derive the formula for the lateral surface area of the cone from the sector?

**Tip:** Always double-check units and simplify calculations step by step for accuracy in geometry problems.

OPAQ is a sector of a circle with centre 0 and radius cm. The angle at the centre is 0. The sector is then formed into a cone by joining the two radii, OP and OQ, together. The radius of the base of the cone is r/3 cm. The cone has a height of 6 cm and a volume of 50π cm³. Find the value of 0.

Solve a geometry problem involving a sector of a circle forming a cone with a base radius of r/3 cm, height of 6 cm, and volume of 50π cm³. Learn to find the angle at the center using key concepts like volume of a cone, arc length, and sector geometry. Suitable for students in Grades 10-12.

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