Tuesday, November 8th

Problem 
Inscribe a circle of radius $r$ in a 30,60,90 triangle. Find the distances from the center of the circle to the vertices as a function of $r$.


Solution
Let $ABC$ be the triangle where $A, B, C$ are the vertices with angles of 30, 90 and 60 degrees respectively. Let $O$ be the center of the circle. Then $OA=r/\sin (15)$, $OB=r/\sin(45)$ and $OC=r/\sin(30)$. It is well known the values of the two later ones, so just remains to calculate the value of $\sin(15)$.


$\sin(15)=\sqrt{\frac{1-\cos(30)}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}$.