how to find slant height using radius and angle for a truncated cone I asked this question a while ago but it got closed. I'm trying to find the slant height or just height for a truncated cone with only the radius and the angles for a hobby project that i've been doing. Its called how to develop, build, and test small liquid-fueled rocket engines by rocket lab. The truncated cone is for the nozzle throat. The base radius (R=1.50225 in) and the top radius (r=0.50075 in). I have looked everywhere but it keeps saying volume and height and surface area. Not learning in 9th grade eather.
      r=0.50079
      ________________ 30˙  
     /                \    |
    /                  \   |
   /                    \  |h= unknown
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 -------------------------- 60˙    looking for slant height or height   R=1.50225

 A: 
See (rather terrible) diagram above.
The half-angle of the cone (from the vertical axis) is $\theta$.
By simple right-angle trigonometry, you know that $l = \frac r{\sin \theta} = \frac r{\cos(90^{\circ} - \theta)}$. I'm giving you both forms because I'm not sure which angle you have.
Similarly, $L =  \frac R{\sin \theta} = \frac R{\cos(90^{\circ} - \theta)}$
Find $L-l$ to get the truncated slant height.
A: Here's a method that I think is especially easy to apply if you just know a little trigonometry.

The diagram is a cross-section of your truncated cone. The vertical line in the middle is the axis (centerline) of the cone.
The other vertical line is parallel to the axis and meets the edge of the circle at the upper end of the truncated cone.
So we now have a right triangle with base $R - r = 1.50225 - 0.50075 = 1.0015.$
The height $h$ is $(R - r) \tan 60^\circ$ and the slant height is 
$(R - r) \sec 60^\circ$.
You can put any other angle in place of $60$ degrees for a different shape of cone.
But the $60$ degree angle is particularly nice because
$\tan 60^\circ = {\sqrt3}$ and $\sec 60^\circ = 2.$
