SOLUTION: A cone of height 9cm has a volume of n cm³ and a curved surface area of n cm², find the vertical angle

Question 1205289: A cone of height 9cm has a volume of n cm³ and a curved surface area of n cm², find the vertical angle
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52851) (Show Source): You can put this solution on YOUR website!
.

Was solved at this forum several years ago. See the link

https://www.algebra.com/algebra/homework/Surface-area/Surface-area.faq.question.1134450.html

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

V = volume of the cone
SA = surface area
LSA = lateral surface area = area of just the curved wall or ceiling
The LSA ignores the area of the flat base.

r = radius = unknown
h = height = 9

Formulas to memorize or have on a reference sheet
V = (1/3)*pi*r^2*h
LSA = pi*r*sqrt(r^2+h^2)

sqrt(r^2+h^2) is the slant height due to the Pythagorean theorem.

Plug h = 9 into each to get
V = 3pi*r^2
LSA = pi*r*sqrt(r^2+81)

Those items are both equal to a variable n (just with different units of course).
That allows us to set them equal to one another to solve for variable r.

V = LSA
3pi*r^2 = pi*r*sqrt(r^2+81)
3pi*r^2 - pi*r*sqrt(r^2+81) = 0
pi*r*(3r - sqrt(r^2+81)) = 0
pi*r = 0 or 3r - sqrt(r^2+81) = 0
r = 0 or 3r - sqrt(r^2+81) = 0

Ignore r = 0 since it's a trivial solution.
We only consider r > 0.

3r - sqrt(r^2+81) = 0
3r = sqrt(r^2+81)
(3r)^2 = (sqrt(r^2+81))^2
9r^2 = r^2+81
9r^2-r^2 = 81
8r^2 = 81
r^2 = 81/8
r = sqrt(81/8)
r = 3.1819805 approximately

The apex angle, or vertex angle, of the cone is the angle at the very top.
Half of this angle is denoted as theta in the diagram.

theta = = greek letter often used for angles.

Focus on one of the right triangles.
tan(angle) = opposite/adjacent
tan(theta) = r/h
tan(theta) = 3.1819805/9
tan(theta) = 0.3535534
theta = arctan(0.3535534)
theta = 19.4712211 degrees approximately

This doubles to 2*19.4712211 = 38.9424422 which is also approximate.
Round that value however needed.

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