Volume of a Cone using Trigonometric Ratios

Quote:

Originally Posted by KelvinScale

Hi everyone,

I can't figure out the following question from my math homework:

A right circular cone is inscribed in a sphere of radius 30 cm. The semi vertical angle of the cone is x. Determine an expression for the volume of the cone in terms of x.

(I can't explain the semi-vertical angle very well, but Imagine the cone is an isocoles triangle, and a line is drawn from the top point down to the base. The angle "x" in this question would be one of the two "angles" formed by the vertical line you drew and the two slanted sides of the triangle).

In case you're wondering, the textbook says that the answer is

V=18000πsin^2(2x)cos^2(x)

This is a rough representation of the picture. Just imagine the angle at the top formed by the height and and slant is "x", and R=30cm.

The biggest right circular cone that can be inscribed in a sphere - Math Central

http://mathcentral.uregina.ca/QQ/dat...strogirl1.html

ok see the picture ny applying since law on the triangle ABD

Attachment 14732

$\frac{e}{\sin (180 - 2x) } = \frac{30}{\sin x}$

note that :- $\sin (180-2x) = \sin 180 \cos 2x - \sin 2x \cos 180 = \sin 2x$

$e = \frac{30 \cdot \sin 2x }{\sin x} = \frac{30\cdot 2\cdot \sin x \cos x}{\sin x} = 60\cos x$

now in the triangle ABC

$\sin x = \frac{R}{e} \Rightarrow R= e\sin x$

$\cos x =\frac{h}{e} \Rightarrow h=e\cos x$

but we find e with respect to x so

$R = 60\sin x\cos x = 30\sin 2x$

$h = 60\cos x \cos x = 60\cos ^2 x$

the volume of the cone given by

$V = \frac{R^2 h \pi}{3}$ h is the height and R is the radius of the base

$V = \frac{(30\sin 2x)^2 (60\cos ^2 x) \pi }{3}$