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

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Therefore the answer is <math>\mathrm{(D)}</math>.
 
Therefore the answer is <math>\mathrm{(D)}</math>.
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==Solution 3==
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Using the previous drawing we know that ACD and DAC are both equal to 20 (40/2 or (180-40)/2). We also know that BAC and BCA are both 70 or (180-40)/2. Thus BDA is 70-20 or 50. Easy. -RealityWrites
  
 
==See also==
 
==See also==

Latest revision as of 23:20, 29 October 2022

The following problem is from both the 2007 AMC 12A #6 and 2007 AMC 10A #8, so both problems redirect to this page.

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 1

2007 AMC12A-6.png

We angle chase and find out that:

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

~minor edits by mobius247

Solution 2

Mihir Borkar Solution 2007 AMC 10A Problem 6 p 2.png

Since triangle $ABC$ is isosceles we know that angle $\angle BAC = \angle BCA$.

Also since triangle $ADC$ is isosceles we know that $\angle DAC = \angle DCA$.

This implies that $\angle BAD = \angle BCD$.

Then the sum of the interior angles of quadrilateral $ABCD$ is $40 + 220 + 2\angle BAD = 360$.

Solving the equation, we get $\angle BAD = 50$.

Therefore the answer is $\mathrm{(D)}$.

Solution 3

Using the previous drawing we know that ACD and DAC are both equal to 20 (40/2 or (180-40)/2). We also know that BAC and BCA are both 70 or (180-40)/2. Thus BDA is 70-20 or 50. Easy. -RealityWrites

See also

2007 AMC 12A (Problems • Answer Key • Resources)
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
2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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. AMC logo.png

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