Lorena Beard

2022-07-09

42 degrees is constructible, how do I show? At the same line, how do I show that 18 degrees has $\mathrm{cos}$ constructible?

Jasper Parsons

Expert

From a regular hexagon, ${60}^{\circ }$ is constructible. From a regular pentagon, ${72}^{\circ }$ is constructible. Hence is the difference ${12}^{\circ }$. Multiply by 4 to find ${48}^{\circ }$, then subtract from ${90}^{\circ }$ to find ${42}^{\circ }$.
Start from a pentagon (${72}^{\circ }$) and half the angle twice to arrive at ${18}^{\circ }$

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