Difference between revisions of "2007 AMC 12A Problems/Problem 6"
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This implies that <math>\angle BAD = \angle BCD</math>. | This implies that <math>\angle BAD = \angle BCD</math>. | ||
− | Then the sum of the angles | + | Then the sum of the interior angles of quadrilateral <math>ABCD</math> is <math>40 + 220 + 2\angle BAD = 360</math>. |
− | Solving the equation we get <math>\angle BAD = 50</math>. | + | Solving the equation, we get <math>\angle BAD = 50</math>. |
Therefore the answer is <math>\mathrm{(D)}</math>. | Therefore the answer is <math>\mathrm{(D)}</math>. | ||
+ | |||
+ | ==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== | ==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 and are isosceles with and . Point is inside triangle , angle measures 40 degrees, and angle measures 140 degrees. What is the degree measure of angle ?
Solution 1
We angle chase and find out that:
~minor edits by mobius247
Solution 2
Since triangle is isosceles we know that angle .
Also since triangle is isosceles we know that .
This implies that .
Then the sum of the interior angles of quadrilateral is .
Solving the equation, we get .
Therefore the answer is .
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.