Hanover, Maryland AA Meetings

  • Please note that AA meeting locations and times tend to change often and quickly, so always check before assuming these times are accurate.
  • Do you have new information about an AA meeting? Please let us know!

Hanover is located in the Annapolis area in Anne Arundel County and Howard County, Maryland. Located south of Baltimore, the unincorporated community has Alcoholics Anonymous programs set up to help members of its community who are struggling with alcohol addiction. If you are one of them, you can find AA meetings in Hanover with a quick search on alcoholicsanonymous.com. The directory organizes information for AA meetings across America, including all AA meetings Maryland. Seeking help from an AA location is one of the most important decisions you can make in your quest to return to sobriety. AA meetings have helped millions of people from all over the world to quit drinking and live fulfilling lives. Various studies have shown that people with an alcohol dependency who take part in AA programs accomplish better end results compared to those who seek professional therapy. One of the primary reasons it has helped so many people is because of its welcoming nature. AA meetings do not discriminate against anyone despite their race, faith, gender, nationality, affiliation, and financial or social status. AA meetings are led by senior members who achieved sobriety through the AA program and have remained sober for many years. These veterans act as mentors, or “sponsors,” who are dedicated to helping younger members and new participants to overcome alcohol addiction.

Where do calls go?

Calls to numbers on a specific treatment center listing will be routed to that treatment center. Additional calls will also be forwarded and returned by one of our treatment partners below.

Calls to any general helpline (non-facility specific 1-8XX numbers) for your visit will be answered by ARK Behavioral Health, a paid advertiser on AlcoholicsAnonymous.com.

All calls are private and confidential.

Who Answers?