A sector of angle 125° is revomed from a thin circular sheet of radius 18cm. it is then folded with straight edges coinciding to form a right circular cone. what are the steps you would use to calculate the base radius, the semi- vertical, and the volume of the cone?

Volume of the cone is 1883.7 cm³

Step-by-step explanation:

The circumference of the full circle with radius 18 cm :

360 := 2*π*18 = 36π cm

125 := 125/360 * 36π

The new circumference is maller:

36π - 125/360 * 36π

36π * 0.652(7)

Calculate the new r based on the new circomference:

2*π * r = 36π * 0.652(7)

r = 36π/2π * 0.652(7)

r = 18 * 0.652(7)

r = 11.75 cm

Based on this radius you can calculate the area of the base of the cone.

area base = π*(11.75)²

The Volume V of this cone = 1/3 π r² * h

You can calculate the height h by using Pythagoras theorum.

The sector is the hypothenusa= 18 cm

The h is the height, which is the "unknown"

The r is the new radius = 11.75 cm

s² = r² + h²

h² = s² - r²

h = √(s² - r²)

h = √(18² - 11.75²)

h = 13.6358901432946 cm

h = 13.636 cm

V cone

V = 1/3 π 11.75² * h

V = 1/3 π 11.75² * √(18² - 11.75²)

V = 1/3 π 11.75² * 13.636

V = 1883.7 cm³

b is 38 degrees.

a is 52 degrees.

this is because if b is 38 and a is 14 degrees more than b than 38+14=52 and 52+38=90 degrees. hope this .

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: