(A) Acetaldehyde forms hemiacetal with $C{H}_{3}OH$ in the presence of dry HCl. 

(R) Hemiacetal contains both alcohol and alkoxy group.

Acetaldehyde when treated with methanol in presence of dry HCl it gives Hemiacetal intermediate but it is not stable so, this on acidification and heating gives final product as 1,1-Diethoxyethane (acetal). Hemiacetal contains both alkoxy and alcohol group. 

Therefore, both (A) and (R) are true and (R) is not the correct explanation of (A).