Question 1100756
 *[illustration des8.JPG].
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So from the diagram,
{{{alpha+beta+90=180}}}
{{{alpha+beta=90}}}
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{{{cos(alpha)=r/CX=r/(20-r)}}}
and
{{{sin(alpha)=20/25=4/5}}}
You also know that,
{{{sin^2(alpha)+cos^2(alpha)=1}}}
{{{16/25+r^2/(20-r)^2=1}}}
{{{r^2/(20-r)^2=9/25}}}
{{{r/(20-r)=3/5}}}
{{{5r=3(20-r)}}}
{{{5r=60-3r}}}
{{{8r=60}}}
Solve for r.
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The volume of the cone is,
{{{V[c]=pi*R^2*h}}}
Calculate R using the Pythagroean theorem,
{{{20^2+R^2=25^2}}}
{{{R=sqrt(625-400)}}}
and the volume of the sphere is,
{{{V[s]=(4/3)pi*r^3}}} with r calculated above.
So the left over volume would be,
{{{V=V[c]-V[s]}}}