AMC8 2014 8, Problem 15

The circumference of the circle with center $O$ is divided into $12$ equal arcs, marked the letters $A$ through $L$ as seen below. What is the number of degrees in the sum of the angles $x$ and $y$? [asy] size(230); defaultpen(linewidth(0.65)); pair O=origin; pair[] circum = new pair[12]; string[] let = {"$A$","$B$","$C$","$D$","$E$","$F$","$G$","$H$","$I$","$J$","$K$","$L$"}; draw(unitcircle); for(int i=0;i<=11;i=i+1) { circum[i]=dir(120-30*i); dot(circum[i],linewidth(2.5)); label(let[i],circum[i],2*dir(circum[i])); } draw(O--circum[4]--circum[0]--circum[6]--circum[8]--cycle); label("$x$",circum[0],2.75*(dir(circum[0]--circum[4])+dir(circum[0]--circum[6]))); label("$y$",circum[6],1.75*(dir(circum[6]--circum[0])+dir(circum[6]--circum[8]))); label("$O$",O,dir(60)); [/asy]

Solution: For this problem, it is useful to know that the measure of an inscribed angle is half the measure of its corresponding central angle. Since each unit arc is $\frac{1}{12}$ of the circle's circumference, each unit central angle measures $\left( \frac{360}{12} \right) ^{\circ}=30^{\circ}$. Then, we know that the central angle of x = $60$, so inscribed angle = $30$. Also, central angle of y = $120$, so inscirbed angle = $60$. Summing both inscribed angles gives $30 + 60 = \fbox{90}.$

Answer: 90