Difference between revisions of "2011 AMC 12B Problems/Problem 10"
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== Solution == | == Solution == | ||
− | Since <math>AB \parallel CD</math>, <math>\angle AMD = \angle CDM</math>, so <math>\angle AMD = \angle CMD = \angle CDM</math>, so <math>\bigtriangleup CMD</math> is isosceles, and hence <math>CM=CD=6</math>. Therefore, <math>\angle BMC = 30^\circ</math>. Therefore <math>\angle AMD=\boxed{\mathrm{(E)}\ 75^\circ}</math> | + | Since <math>AB \parallel CD</math>, <math>\angle AMD = \angle CDM</math>, so <math>\angle AMD = \angle CMD = \angle CDM</math>, so <math>\bigtriangleup CMD</math> is isosceles, and hence <math>CM=CD=6</math>. Therefore, <math>\triangle BMC</math> is a 30-60-90 triangle with <math>\angle BMC = 30^\circ</math>. Therefore <math>\angle AMD=\boxed{\mathrm{(E)}\ 75^\circ}</math> |
== See also == | == See also == | ||
{{AMC12 box|year=2011|ab=B|num-b=9|num-a=11}} | {{AMC12 box|year=2011|ab=B|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:39, 20 June 2024
Problem
Rectangle has and . Point is chosen on side so that . What is the degree measure of ?
Solution
Since , , so , so is isosceles, and hence . Therefore, is a 30-60-90 triangle with . Therefore
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 |
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