Let $ M $ be a point in the $ ABCD $ rectangle, such that $ \angle AMD = 90 ^ {\circ}, BC = 2MD $ and $ CD = AM $. Find the value of $16 \cos^2 \angle CMD$
Attemp: $ AD = BC = 2 MD $, which implies that $\angle MDA = 30 ^ {\circ}$, and $AM = \sqrt {3} \cdot MD = CD$. Then use law of cosine to find $MC$, and then use law of sine to find $\sin \angle CMD$,
Correct?
Yes, it looks good. For the purpose of the problem we can rescale the rectangle, so that $AD=2$. Then $M$ is a point on the circle with diameter $AD$ on the one side, and a point on the circle centered in $D$ and radius $1$ on the other side. The triangle $\Delta AMD$ has a right angle in $M$, $60^\circ$ in $D$ (and $30^\circ$ in $A$, which we do not need, there is a typo in the OP regarding these angles...) so $AM=AB=CD=\sqrt 3$.
We work now in the triangle $\Delta DMC$, it has $30^\circ$ in $D$, so $$ MC^2 = 1^2+\sqrt 3^2-2\cdot 1\cdot \sqrt 3\;\cos 30^\circ =1=1^2 \ . $$
So $\Delta MCD$ is isosceles in $M$, and the angle in $M$ is $120^\circ$.
(At least now we have a better way to show the above. Let $M'$ be the projection of $M$ on $CD$, then $MD=1$, $\angle MDM'=30^\circ$ gives $DM'=\sqrt 3/2$, so $M'$ is the mid point of $DC$, and so on...)
From here and $\cos 120^\circ = -1/2$ things have a quick end.