If $\cos 60^{\circ}=\cos 45^{\circ} \cos \theta$ with $0^{\circ} \leq \theta \leq 90^{\circ}$, then $\theta$ equals
(A) $0^{\circ}$
(B) $15^{\circ}$
(C) $30^{\circ}$
(D) $45^{\circ}$
(E) $60^{\circ}$
Since $\cos 60^{\circ}=\frac{1}{2}$ and $\cos 45^{\circ}=\frac{1}{\sqrt{2}}$, then the given equation $\cos 60^{\circ}=\cos 45^{\circ} \cos \theta$ becomes $\frac{1}{2}=\frac{1}{\sqrt{2}} \cos \theta$.

Therefore, $\cos \theta=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$.

Since $0^{\circ} \leq \theta \leq 90^{\circ}$, then $\theta=45^{\circ}$.

ANSWER: (D)