Let AC be X, and use sine law in triangle CBD and triangle ADB at the same time, and two formulas can be obtained, one is BC/SIND = 1/SIN30, and the other is (X+1)/SIN120 =1/SIND.

I use the second formula to express sinD alone: SIND = SIN120/(X+1), and then bring this SIND into the first formula. Then solve this seemingly complicated equation, which is simple after factorization, and get (x+2) (x 3-2) = 0. Because AC cannot be negative, X 3 = 2. x= 1.25992 105