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Difference between revisions of "2003 AIME I Problems/Problem 10"

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Clearing [[denominator]]s, evaluating <math>\sin 150^\circ = \frac 12</math> and applying one of our [[trigonometric identities]] to the result gives
 
Clearing [[denominator]]s, evaluating <math>\sin 150^\circ = \frac 12</math> and applying one of our [[trigonometric identities]] to the result gives
  
−
<math>\frac{1}{2} \cos 7^\circ - \theta = \sin 7^\circ \sin \theta</math>
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<math>\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta</math>
  
 
and multiplying through by 2 and applying the [[double angle formula]] gives
 
and multiplying through by 2 and applying the [[double angle formula]] gives

Revision as of 14:20, 12 March 2007

Problem

Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ.$ Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ.$ Find the number of degrees in $\angle CMB.$

Solution

From the givens, we have the following angle measures: $m\angle AMC = 150^\circ$, $m\angle MCB = 83^\circ$. If we define $m\angle CMB = \theta$ then we also have $m\angle CBM = 97^\circ - \theta$. Then Apply the Law of Sines to triangles $\triangle AMC$ and $\triangle BMC$ to get

$\frac{\sin 150^\circ}{\sin 7^\circ} = \frac{AC}{CM} = \frac{BC}{CM} = \frac{\sin \theta}{\sin 97^\circ - \theta}$

Clearing denominators, evaluating $\sin 150^\circ = \frac 12$ and applying one of our trigonometric identities to the result gives

$\frac{1}{2} \cos (7^\circ - \theta )= \sin 7^\circ \sin \theta$

and multiplying through by 2 and applying the double angle formula gives

$\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta$

and so $\cos 7^\circ \cos \theta = \sin 7^\circ \sin\theta$

and, since $0^\circ < \theta < 180^\circ$, we must have $\theta = 83^\circ$, so the answer is $083$.

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
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