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

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== Solution ==
 
== Solution ==
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{{solution}}
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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
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<math>\frac{\sin 150^\circ}{\sin 7^\circ} = \frac{AC}{CM} = \frac{BC}{CM} = \frac{\sin \theta}{\sin 97^\circ - \theta}</math>
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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
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<math>\frac{1}{2} \cos 7^\circ - \theta = \sin 7^\circ \sin \theta</math>
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and multiplying through by 2 and applying the [[double angle formula]] gives
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<math>\cos 7^\circ\cos\theta + \sin7^\circ\sin\theta = 2 \sin7^\circ \sin\theta</math>
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and so <math>\cos 7^\circ \cos \theta = \sin 7^\circ \sin\theta</math>
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and, since <math>0^\circ < \theta < 180^circ</math>, we must have <math>\theta = 83^\circ</math>, so the answer is <math>083</math>.
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== See also ==
 
== See also ==
 
* [[2003 AIME I Problems/Problem 9 | Previous problem]]
 
* [[2003 AIME I Problems/Problem 9 | Previous problem]]
 
* [[2003 AIME I Problems/Problem 11 | Next problem]]
 
* [[2003 AIME I Problems/Problem 11 | Next problem]]
 
* [[2003 AIME I Problems]]
 
* [[2003 AIME I Problems]]
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[[Category:Intermediate Geometry Problems]]
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[[Category:Intermediate Trigonometry Problems]]

Revision as of 17:05, 24 October 2006

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

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