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
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.