Nick,
I drew a diagram of and labeled some points.
The "flat pattern" will be part of a circle of radius PS so I need that length. I know that TS = 24" so I need to find PT. The triangles PQT and PRS are similar so
^{PS}/_{28.5} = ^{PT}/_{11.5} or ^{(PT+ TS)}/_{28.5} = ^{PT}/_{11.5}
and hence
^{(PT+ 24)}/_{28.5} = ^{PT}/_{11.5}
I solved for PT and got PT = 16.24", and thus PS = 24 + 14.24 = 40.24".
I can now see that the "flat pattern" is part of a sector of a circle of radius 40.24".
All that remains is to determine the measure of the angle SPV.
The relationship I need to use here is
a = r theta
where a is the length of the arc SV, r is the radius PS and theta is the measure of the angle SPV in radians. The length of the arc SV is the circumference of the base of the cone. The base of the cone is a circle of radius 57" so
a = 2 radius = 2 28.5" = 179.07 "
Thus
a = r theta gives 179.07 = 40.24 theta
This gives theta = 4.45 radians. There are radians in 180^{o} so 4.45 radians is
^{180}/_{} 4.45 = 254.0^{o}
Draw two concentric circles of radius 40.24" and 16.24". Remove a sector with centre angle of 360^{0} 254^{o} = 106^{o} and roll up what remains to form the truncated cone.
Penny
