A sector of a circle of radius 7.2 cm substance and angle of 300 degree at the centre it is used to form a cone fine the vertical angle of the cone

Answer 1

The circumference of the circle is 2πr = 2π(7.2) = 14.4π cm.

The arc length of the sector is 300/360 times the circumference, which is (300/360)(14.4π) = 12π cm.

This arc length is the base circumference of the cone. Let the height of the cone be h and the slant height be s.

Then the vertical angle of the cone is equal to 2 times the inverse tangent of (s/2h).

To find s and h, we can use the formulas for the volume and surface area of a cone:

V = (1/3)πr^2h = (1/3)π(7.2)^2h and S = πrs + πr^2 = πs(7.2 + s).

Using the arc length formula, we know that the base circumference of the cone is 12π cm. Therefore,

2πr = 12π

r = 6 cm

Substituting this value into the formulas for V and S, we get:

V = (1/3)π(6)^2h = 12πh

S = πs(6 + s)

Since the slant height s is equal to the square root of (h^2 + r^2), we can substitute this in and simplify:

S = π√(h^2 + 6^2)(6 + √(h^2 + 6^2))

Now we can set up an equation for the volume and surface area:

12πh = (1/3)π(6)^2h

πs(6 + s) = 12π

Simplifying these equations:

4h = 36

s(6 + s) = 12

We can solve for h and s:

h = 9 cm

s = 1 cm

Substituting these values into the formula for the vertical angle of the cone:

2tan^-1(s/2h) = 2tan^-1(1/18)

The vertical angle of the cone is approximately 6.1 degrees.