Difference between revisions of "2011 AMC 10B Problems/Problem 17"
(→Problem) |
|||
Line 33: | Line 33: | ||
<math> \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140</math> | <math> \textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140</math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
== Solution == | == Solution == |
Revision as of 11:43, 13 August 2014
Problem
In the given circle, the diameter is parallel to
, and
is parallel to
. The angles
and
are in the ratio
. What is the degree measure of angle
?
![[asy] unitsize(7mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; real r=3; pair A=(-3cos(80),-3sin(80)); pair D=(3cos(80),3sin(80)), C=(-3cos(80),3sin(80)); pair O=(0,0), E=(-3,0), B=(3,0); path outer=Circle(O,r); draw(outer); draw(E--B); draw(E--A); draw(B--A); draw(E--D); draw(C--D); draw(B--C); pair[] ps={A,B,C,D,E,O}; dot(ps); label("$A$",A,N); label("$B$",B,NE); label("$C$",C,S); label("$D$",D,S); label("$E$",E,NW); label("$$",O,N); [/asy]](http://latex.artofproblemsolving.com/b/a/d/bad25a60446625a8eac18a0209ed3ab5d7c02eff.png)
Solution
We can let be
and
be
because they are in the ratio
. When an inscribed angle contains the diameter, the inscribed angle is a right angle. Therefore by triangle sum theorem,
and
.
because they are alternate interior angles and
. Opposite angles in a cyclic quadrilateral are supplementary, so
. Use substitution to get
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.