Difference between revisions of "2011 AMC 10B Problems/Problem 17"
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− | == Solution == | + | == Solution 1== |
We can let <math>\angle AEB</math> be <math>4x</math> and <math>\angle ABE</math> be <math>5x</math> because they are in the ratio <math>4 : 5</math>. When an [[inscribed angle]] contains the [[diameter]], the inscribed angle is a [[right angle]]. Therefore by triangle sum theorem, <math>4x+5x+90=180 \longrightarrow x=10</math> and <math>\angle ABE = 50</math>. | We can let <math>\angle AEB</math> be <math>4x</math> and <math>\angle ABE</math> be <math>5x</math> because they are in the ratio <math>4 : 5</math>. When an [[inscribed angle]] contains the [[diameter]], the inscribed angle is a [[right angle]]. Therefore by triangle sum theorem, <math>4x+5x+90=180 \longrightarrow x=10</math> and <math>\angle ABE = 50</math>. | ||
<math>\angle ABE = \angle BED</math> because they are alternate interior angles and <math>\overline{AB} \parallel \overline{ED}</math>. Opposite angles in a [[cyclic]] quadrilateral are [[supplementary]], so <math>\angle BED + \angle BCD = 180</math>. Use substitution to get <math>\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{\textbf{(C)} 130}</math> | <math>\angle ABE = \angle BED</math> because they are alternate interior angles and <math>\overline{AB} \parallel \overline{ED}</math>. Opposite angles in a [[cyclic]] quadrilateral are [[supplementary]], so <math>\angle BED + \angle BCD = 180</math>. Use substitution to get <math>\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{\textbf{(C)} 130}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | Note <math>\angle ABE = \angle BED=50</math> as before. The sum of the interior angles for quadrilateral <math>EBCD</math> is <math>360</math>. Denote the center of the circle as <math>P</math>. <math>\angle PDE = \angle PED = 50</math>. Denote <math>\angle PDC = \angle PCD = x</math> and <math>\angle PBC = \angle PCB = y</math>. We wish to find <math>\angle BCD = x+y</math>. Our equation is <math>(\angle PDE +\angle PED)+(\angle PDC+\angle PCD)+(\angle PBC + \angle PCB) = 360 \longrightarrow 2(50) + 2x +2y = 360</math>. Our final equation becomes <math>2(x+y)+100 = 360</math>. After subtracting <math>100</math> and dividing by <math>2</math>, our answer becomes <math>x+y=\boxed{\textbf{(C)} 130}</math> | ||
== See Also== | == See Also== |
Revision as of 20:55, 24 December 2017
Contents
[hide]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 1
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
Solution 2
Note as before. The sum of the interior angles for quadrilateral
is
. Denote the center of the circle as
.
. Denote
and
. We wish to find
. Our equation is
. Our final equation becomes
. After subtracting
and dividing by
, our answer becomes
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 |
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