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OPAQ is a sector of a circle with centre 0 and radius r 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 3 cm. The cone has a height of 6 cm and a volume of 50 cm'. Find the value of 0.

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

OPAQ is a sector of a circle with center O and radius r cm. The angle at the center is θ. 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 3 cm. The cone has a height of 6 cm and a volume of 50 cm³. Find the value of θ.

This problem involves calculating the value of the angle θ when a sector of a circle is formed into a cone. Given the base radius of 3 cm, height 6 cm, and a cone volume of 50 cm³, the problem explores key geometric concepts and mensuration formulas suitable for high school students.

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