7・12 If $\theta$ is an acute angle, $\sin 2 \theta=a$, then $\sin \theta+\cos \theta$ equals
(A) $\sqrt{a+1}$.
(B) $(\sqrt{2}-1) a+1$.
(C) $\sqrt{a+1}-\sqrt{a^{2}-a}$.
(D) $\sqrt{a+1}+\sqrt{a^{2}-a}$.
(E) $\sqrt{a+1}+a^{2}-a$.
(27th American High School Mathematics Examination, 1976)
[Solution]
$$
\text { Given } \begin{aligned}
(\sin \theta+\cos \theta)^{2} & =\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta \\
& =1+\sin 2 \theta=1+a,
\end{aligned}
$$

Since $\theta$ is an acute angle,
$$
\therefore \quad \sin \theta+\cos \theta=\sqrt{1+a} \text {. }
$$

Therefore, the correct choice is $(A)$.