



 
Hi Mary, I drew a diagram of your flat top cone (truncated cone), extended it to a full cone, and added some labels. My diagram is not to scale. $BC = 12.5 \mbox{ cm, } FD = 7 \mbox{ cm, and }ED = 15 cm.$ Hence $EC = 5.5 \mbox{ cm.}$ Triangle $DEC$ is a right triangle so Pythagoras' Theorem gives us \[CD^2 = DE^2 + EC^2 = 225 + 30.25 = 255.25 \mbox{ so } CD = 16.89 \mbox{ cm.}\] Triangles $ABC$ and $DEC$ are similar and hence \[\frac{CA}{BC} = \frac{CD}{EC} \mbox{ and hence } CA = 38.39 \mbox{ cm.}\] Thus $AD = 38.39  16.89 = 21.50 \mbox{ cm.}$ Imagine that you slice the surface of the cone along the line $CA$ and roll it out flat to form a sector of a circle. There are two copies of $C$ on the rolled out image, and I called one of them $C^{\prime}.$ The length of the arc from $C$ to $C^\prime$ is the circumference of the circular base of the cone so the length is $2 \times \pi \ 12.5$ cm. This arc is a fraction of the circumference of the circle with centre $A$ and radius $AC.$ Likewise the measure of the angle $t^o$ at the center of the sector is a fraction of $360^{0},$ the angle all the way around the circle. By the symmetry of a circle these fractions are the same. Thus \[\frac{2 \times \pi \ 12.5}{2 \times \pi \ 38.39} = \frac{t^o}{360^o}\] and hence \[t^o = 117.2^{o}.\] I hope this helps, 



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