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3 A cone of base radius 7 cm was made from a sector of a circle which subtends an angle of \( 320^{\circ} \) at the centre. Find the radius of the circle and the vertical angle of the cone.

Ask by Reese Zimmerman. in Nigeria
Mar 10,2025

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The radius of the circle is \( \frac{63}{8} \) cm, and the vertical angle of the cone is \( 2\arcsin\left(\frac{8}{9}\right) \).

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Bonus Knowledge

To find the radius of the original circle from which the cone was made, we can use the relationship between the arc length of the sector and the circumference of the base of the cone. The circumference of the cone's base is calculated using \( C = 2\pi r \) where \( r \) is the radius of the cone's base. For the cone, with a base radius of 7 cm, the circumference \( C = 2\pi \times 7 = 14\pi \) cm. The arc length of the sector that forms the cone is given by the formula \( L = \frac{\theta}{360^\circ} \times 2\pi R \), where \( R \) is the radius of the original circle and \( \theta \) is the angle in degrees. Using \( \theta = 320^\circ \), we have: \[ 14\pi = \frac{320}{360} \times 2\pi R \] Simplifying this equation, we get: \[ 14 = \frac{32}{36} \times R \quad ( \text{we can divide by } 2\pi) \] \[ 14 = \frac{8}{9} R \] Thus, \[ R = 14 \times \frac{9}{8} = 15.75 \text{ cm} \] So, the radius of the circle is **15.75 cm**. Next, to find the vertical angle of the cone, let \( r = 7 \) cm (base radius) and \( R = 15.75 \) cm (slant height of the cone). The vertical angle \( \alpha \) can be found using trigonometry: The relationship is given by: \[ \tan\left(\frac{\alpha}{2}\right) = \frac{r}{h} \] Where \( h \) is the height of the cone, which we can derive using the Pythagorean theorem. First, we find the height \( h \): \[ h = \sqrt{R^2 - r^2} \] Calculating: \[ h = \sqrt{(15.75)^2 - (7)^2} = \sqrt{248.0625 - 49} = \sqrt{199.0625} \approx 14.1 \text{ cm} \] Now, using \( r = 7 \) cm and \( h \approx 14.1 \) cm: \[ \tan\left(\frac{\alpha}{2}\right) = \frac{7}{14.1} \approx 0.496 \] Taking the arctangent: \[ \frac{\alpha}{2} \approx \tan^{-1}(0.496) \approx 26.3^\circ \] \[ \alpha \approx 52.6^\circ \] Thus, the vertical angle of the cone is approximately **52.6 degrees**.

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