Difference between revisions of "2003 AIME I Problems/Problem 10"
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From the givens, we have the following [[angle]] [[measure]]s: <math>m\angle AMC = 150^\circ</math>, <math>m\angle MCB = 83^\circ</math>. If we define <math>m\angle CMB = \theta</math> then we also have <math>m\angle CBM = 97^\circ - \theta</math>. Then Apply the [[Law of Sines]] to triangles <math>\triangle AMC</math> and <math>\triangle BMC</math> to get | From the givens, we have the following [[angle]] [[measure]]s: <math>m\angle AMC = 150^\circ</math>, <math>m\angle MCB = 83^\circ</math>. If we define <math>m\angle CMB = \theta</math> then we also have <math>m\angle CBM = 97^\circ - \theta</math>. Then Apply the [[Law of Sines]] to triangles <math>\triangle AMC</math> and <math>\triangle BMC</math> to get | ||
Revision as of 15:25, 16 September 2007
Problem
Triangle is isosceles with and Point is in the interior of the triangle so that and Find the number of degrees in
Solution
From the givens, we have the following angle measures: , . If we define then we also have . Then Apply the Law of Sines to triangles and to get
Clearing denominators, evaluating and applying one of our trigonometric identities to the result gives
and multiplying through by 2 and applying the double angle formula gives
and so
and, since , we must have , so the answer is .
See also
2003 AIME I (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 | ||
All AIME Problems and Solutions |