problem on spheres

Draw a line from  the  top of the  cone perpendicular to  its  base.....this is the  height  of  the  cone = CF =  h

This forms a  right triangle  with one leg = FB =  8   and the  hypotenuse = BC = sqrt (h^2 + 8^2) = sqrt (h^2 +64)

Next draw a line from the  center of the  sphere perpendicular to  the  side  of the  cone

This forms a right triangle similar to  the  first  one with a leg DE =  4 and  a  hypotenuse=  DC  = h - 4

By similar triangles  

DE / DC   =  FB / BC

4 / (h - 4)  =  8 / sqrt ( h^2 + 64)  cross-multiply

4sqrt (h^2 + 64)  =  8 (h-4)   

sqrt ( h^2 + 64)  = 2(h - 4)   square both sides

h^2 +  64 =  4 ( h^2 - 8h + 16)

h^2 + 64  =  4h^2 - 32h + 64

3h^2 - 32h =  0

h ( 3h - 32) = 0

Setting the second factor =  0

3h - 32  = 0

3h = 32

h = 32/ 3 in  ≈  10.7 in

What is the height of the cone?

Hello bingboy!

$tan\frac{\propto}{2}=\frac{4}{8}=0.5\\ \propto\ =2\ arctan\ 0.5=53.13^\circ\\ {\color{blue}m=}-\ tan\propto\ =\color{blue}-\frac{4}{3}$

P(8,0)

$y=m(x-x_P)+y_P\\ y=-\frac{4}{3}(x-8)+0\\ y=-\frac{4}{3}x+\frac{32}{3}\\ \color{blue}y=-\frac{4}{3}x+10.\overline{6}$

The height of the cone is 10.67 inches.

  !