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Difference between revisions of "2014 AMC 8 Problems/Problem 15"

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<math> \textbf{(A) }75\qquad\textbf{(B) }80\qquad\textbf{(C) }90\qquad\textbf{(D) }120\qquad\textbf{(E) }150 </math>
 
<math> \textbf{(A) }75\qquad\textbf{(B) }80\qquad\textbf{(C) }90\qquad\textbf{(D) }120\qquad\textbf{(E) }150 </math>
 
==Solution==
 
==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 <math>\frac{1}{12}</math> of the circle's circumference, each unit central angle measures <math>\left( \frac{360}{12} \right) ^{\circ}=30^{\circ}</math>. Then, we know that the inscribed arc of <math>\angle x=60^{\circ}</math> so <math>m\angle x=30^{\circ}</math>; and the inscribed arc of <math>\angle y=120^{\circ}</math> so <math>m\angle y=60^{\circ}</math>. <math>m\angle x+m\angle y=30+60=\framebox{(C) 90}</math>.
+
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 <math>\frac{1}{12}</math> of the circle's circumference, each unit central angle measures <math>\left( \frac{360}{12} \right) ^{\circ}=30^{\circ}</math>. Then, we know that the central angle of <math>\x=60^{\circ}</math> so <math>m\angle x=30^{\circ}</math>; and the central angle of <math>\ y=120^{\circ}</math> so <math>m\angle y=60^{\circ}</math>. <math>m\angle x+m\angle y=30+60=\framebox{(C) 90}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2014|num-b=14|num-a=16}}
 
{{AMC8 box|year=2014|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:06, 10 July 2021

Problem

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]

$\textbf{(A) }75\qquad\textbf{(B) }80\qquad\textbf{(C) }90\qquad\textbf{(D) }120\qquad\textbf{(E) }150$

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 $\x=60^{\circ}$ (Error compiling LaTeX. Unknown error_msg) so $m\angle x=30^{\circ}$; and the central angle of $\ y=120^{\circ}$ so $m\angle y=60^{\circ}$. $m\angle x+m\angle y=30+60=\framebox{(C) 90}$.

See Also

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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