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

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==Solution==
 
==Solution==
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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 x = 60, so inscribed angle = 30. Also, central angle of y = 120, so inscirbed angle = 60. Summing both inscribed angles gives 30 + 60 = 90.
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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 x = 60, so inscribed angle = 30. Also, central angle of y = 120, so inscirbed angle = <math>60</math>. Summing both inscribed angles gives <math>30 + 60 = 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 15:45, 11 November 2022

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]

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

Video Solution

https://www.youtube.com/watch?v=qseG63LK4AU

https://youtu.be/aZhjhb3mMfg ~savannahsolver

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 $30 + 60 = 90.$

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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