



 
Hi Barbie, I drew a diagram of the example you sent (not to scale) as I understand it. Let me know if I have something incorrect. You are correct, this is a frustum of a cone. I am going to extend the diagram to form the full cone and a some labels. I wasn't sure which angle you were looking for but from a page on Wikepedia I think it is the angle $BCA.$ Let the distance from $G$ to $C$ be $x$ mm. Triangles $BCA$ and $FCE$ are similar so \[\frac{AB}{DC} = \frac{EF}{GC} \mbox{ or for your example } \frac{3.65}{1.2 + x} = \frac{1.98}{x}.\] Simplifying the expression for the example gives \[3.65 \times x = 1.98 \times {(1.2 + x)}\] or \[(3.65 \times x = 2.376 + 1.98 \times x.\] This gives $x = 1.4228$ mm. But the tangent of the angle $DCA$ is $\large \frac{AB/2}{DC}.$ For your example this gives \[\tan(DCA) = \frac{3.65/2}{1.20 + 1.4228} = 0.6958\] and hence the measure of the angle $DCA$ is \[\tan^{1}(0.6958) = 34.8^{o}\] and the measure of angle $BCA$ is $2 \times 34.8 = 69.6^{o}.$ You were expecting $90^{o}.$ Is my diagram incorrect? do you see an error in my calculations? Hariey




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