2003 AIME I Problems/Problem 10

Revision as of 17:34, 8 March 2007 by Azjps (talk | contribs) (See also: box)

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 (ProblemsAnswer KeyResources)
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