Difference between revisions of "2000 AMC 8 Problems/Problem 24"

(Created page with "If <math> \angle A = 20^\circ </math> and <math> \angle AFG =\angle AGF </math>, then <math> \angle B+\angle D = </math> <asy> pair A,B,C,D,EE,F,G; A = (0,0); B = (9,4); C = (21...")
 
(Solution + see also box)
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==Problem==
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If <math> \angle A = 20^\circ </math> and <math> \angle AFG =\angle AGF </math>, then <math> \angle B+\angle D = </math>
 
If <math> \angle A = 20^\circ </math> and <math> \angle AFG =\angle AGF </math>, then <math> \angle B+\angle D = </math>
  
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<math> \text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ </math>
 
<math> \text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ </math>
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==Solution==
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As a strategy, think of how <math>\angle B + \angle D</math> would be determined, particularly without determining either of the angles individually, since it may not be possible to determine <math>\angle B</math> or <math>\angle D</math> alone.  If you see <math>\triangle BFD</math>, the you can see that the problem is solved quickly after determining <math>\angle BFD</math>.
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But start with <math>\triangle AGF</math>, since that's where most of our information is.  Looking at <math>\triangle AGF</math>, since <math>\angle F = \angle G</math>, and <math>\angle A = 20</math>, we can write:
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<math>\angle A + \angle G + \angle F = 180</math>
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<math>20 + 2\angle F = 180</math>
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<math>\angle AFG = 80</math>
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By noting that <math>\angle AFG</math> and <math>\angle GFD</math> make a striaght line, we know
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<math>\angle AFG + \angle GFD = 180</math>
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<math>80 + \angle GFD = 180</math>
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<math>\angle GFD = 100</math>
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Ignoring all other parts of the figure and looking only at <math>\triangle BFD</math>, you see that <math>\angle B + \angle D + \angle F = 180</math>.  But <math>\angle F</math> is the same as <math>\angle GFD</math>.  Therefore:
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<math>\angle B + \angle D + \angle GFD = 180</math>
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<math>\angle B + \angle D + 100 = 180</math>
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<math>\angle B + \angle D = 80^\circ</math>, and the answer is thus <math>\boxed{D}</math>
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==See Also==
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{{AMC8 box|year=2000|num-b=23|num-a=25}}

Revision as of 22:09, 30 July 2011

Problem

If $\angle A = 20^\circ$ and $\angle AFG =\angle AGF$, then $\angle B+\angle D =$

[asy] pair A,B,C,D,EE,F,G; A = (0,0); B = (9,4); C = (21,0); D = (13,-12); EE = (4,-16); F = (13/2,-6); G = (8,0); draw(A--C--EE--B--D--cycle); label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,WSW); label("$G$",G,NW);[/asy]

$\text{(A)}\ 48^\circ\qquad\text{(B)}\ 60^\circ\qquad\text{(C)}\ 72^\circ\qquad\text{(D)}\ 80^\circ\qquad\text{(E)}\ 90^\circ$

Solution

As a strategy, think of how $\angle B + \angle D$ would be determined, particularly without determining either of the angles individually, since it may not be possible to determine $\angle B$ or $\angle D$ alone. If you see $\triangle BFD$, the you can see that the problem is solved quickly after determining $\angle BFD$.

But start with $\triangle AGF$, since that's where most of our information is. Looking at $\triangle AGF$, since $\angle F = \angle G$, and $\angle A = 20$, we can write:

$\angle A + \angle G + \angle F = 180$

$20 + 2\angle F = 180$

$\angle AFG = 80$

By noting that $\angle AFG$ and $\angle GFD$ make a striaght line, we know

$\angle AFG + \angle GFD = 180$

$80 + \angle GFD = 180$

$\angle GFD = 100$

Ignoring all other parts of the figure and looking only at $\triangle BFD$, you see that $\angle B + \angle D + \angle F = 180$. But $\angle F$ is the same as $\angle GFD$. Therefore:

$\angle B + \angle D + \angle GFD = 180$ $\angle B + \angle D + 100 = 180$ $\angle B + \angle D = 80^\circ$, and the answer is thus $\boxed{D}$


See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AJHSME/AMC 8 Problems and Solutions
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