Sure, let's solve the problem step by step:
### Given Data:
- Radius of the circle = 7 cm
- Angle subtended by the sector at the center of the circle = 210°
### Forming the Cone:
When the sector is formed into a cone, the arc of the sector becomes the circumference of the base of the cone, and the radius of the original circle becomes the slant height of the cone.
#### Step-by-Step Solution:
#### (a) Base Radius of the Cone
The arc length of the sector is calculated as follows:
[tex]\[ \text{Arc Length} = \theta \times r \][/tex]
where [tex]\(\theta\)[/tex] is in radians and [tex]\(r\)[/tex] is the radius.
First, convert the angle to radians:
[tex]\[ \theta = 210^\circ \times \frac{\pi}{180^\circ} = \frac{7\pi}{6} \text{ radians} \][/tex]
Now, calculate the arc length:
[tex]\[ \text{Arc Length} = \frac{7\pi}{6} \times 7 \, \text{cm} = \frac{49\pi}{6} \, \text{cm} \][/tex]
The arc length becomes the circumference of the base of the cone. So, we set the circumference of the base of the cone equal to the arc length:
[tex]\[ 2\pi \times \text{Base Radius} = \frac{49\pi}{6} \][/tex]
Solving for the base radius:
[tex]\[ \text{Base Radius} = \frac{49\pi}{6} \div 2\pi = \frac{49}{12} \approx 4.083 \, \text{cm} \][/tex]
#### (b) Height of the Cone
We use the Pythagorean theorem to find the height of the cone. The slant height [tex]\(l\)[/tex] is the radius of the original circle (7 cm) and the base radius we have calculated.
[tex]\[ l^2 = \text{Base Radius}^2 + \text{Height}^2 \][/tex]
[tex]\[ 7^2 = 4.083^2 + \text{Height}^2 \][/tex]
[tex]\[ 49 = 16.67 + \text{Height}^2 \][/tex]
[tex]\[ \text{Height}^2 = 49 - 16.67 \approx 32.33 \][/tex]
[tex]\[ \text{Height} = \sqrt{32.33} \approx 5.686 \, \text{cm} \][/tex]
#### (c) Volume of the Cone
The volume [tex]\(V\)[/tex] of the cone is calculated using the formula:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
where [tex]\(r\)[/tex] is the base radius and [tex]\(h\)[/tex] is the height.
[tex]\[ V = \frac{1}{3} \pi (4.083)^2 \times 5.686 \][/tex]
[tex]\[ V \approx \frac{1}{3} \pi \times 16.67 \times 5.686 \][/tex]
[tex]\[ V \approx \frac{1}{3} \pi \times 94.84 \][/tex]
[tex]\[ V \approx 99.27 \, \text{cm}^3 \][/tex]
#### (d) Vertical Angle of the Cone
The vertical angle [tex]\(\alpha\)[/tex] of the cone can be found using the relationship:
[tex]\[ \tan{\frac{\alpha}{2}} = \frac{\text{Base Radius}}{\text{Height}} \][/tex]
[tex]\[ \frac{\alpha}{2} = \tan^{-1}{\left(\frac{4.083}{5.686}\right)} \][/tex]
[tex]\[ \frac{\alpha}{2} \approx \tan^{-1}{0.718} \][/tex]
[tex]\[ \frac{\alpha}{2} \approx 35.685^\circ \][/tex]
[tex]\[ \alpha \approx 2 \times 35.685^\circ = 71.37^\circ \][/tex]
### Summary of Results:
1. Base Radius of the Cone: [tex]\(4.083 \, \text{cm}\)[/tex]
2. Height of the Cone: [tex]\(5.686 \, \text{cm}\)[/tex]
3. Volume of the Cone: [tex]\(99.27 \, \text{cm}^3\)[/tex]
4. Vertical Angle of the Cone: [tex]\(71.37^\circ\)[/tex]
