Difference between revisions of "2007 AMC 12A Problems/Problem 6"

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==Problem==
 
==Problem==
Triangles <math>ABC</math> and <math>ADC</math> are [[isosceles]] with <math>AB=BC</math> and <math>\displaystyle AD=DC</math>. Point <math>D</math> is inside triangle <math>ABC</math>, angle <math>ABC</math> measures 40 degrees, and angle <math>ADC</math> measures 140 degrees. What is the degree measure of angle <math>BAD</math>?
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Triangles <math>ABC</math> and <math>ADC</math> are [[isosceles]] with <math>AB=BC</math> and <math>AD=DC</math>. Point <math>D</math> is inside triangle <math>ABC</math>, angle <math>ABC</math> measures 40 degrees, and angle <math>ADC</math> measures 140 degrees. What is the degree measure of angle <math>BAD</math>?
  
 
<math>\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 40\qquad \mathrm{(D)}\ 50\qquad \mathrm{(E)}\ 60</math>
 
<math>\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 40\qquad \mathrm{(D)}\ 50\qquad \mathrm{(E)}\ 60</math>

Revision as of 18:49, 28 September 2007

Problem

Triangles $ABC$ and $ADC$ are isosceles with $AB=BC$ and $AD=DC$. Point $D$ is inside triangle $ABC$, angle $ABC$ measures 40 degrees, and angle $ADC$ measures 140 degrees. What is the degree measure of angle $BAD$?

$\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 30\qquad \mathrm{(C)}\ 40\qquad \mathrm{(D)}\ 50\qquad \mathrm{(E)}\ 60$

Solution

2007 AMC12A-6.png

We angle chase, and find out that:

  • $DAC=\frac{180-140}{2} = 20$
  • $BAC=\frac{180-40}{2} = 70$
  • $BAD=BAC-DAC=50\ \mathrm{(A)}$

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 12 Problems and Solutions