Ответ:

triangle ABC , AB = 2 , angle A = 30 deg , angle B = 45 deg

\textless ACB = 180 ^ c

(AB)/(sin ACB) = (AC)/(sin ABC)

AC = (2sin 45 deg)/(sin(90 deg + 15 deg)) = 2/c_{0}

= (sqrt(2))/(cos 45 deg * cos 30 deg + sin 45 deg * sin 30 deg) =

= (4sqrt(2))/(sqrt(2) * (sqrt(3) + 1)) = 4/(sqrt(3) + 1) = (4(sqrt(3) - 1))/((sqrt(3) + 1)(sqrt(3) - 1)) =

= (4(sqrt(3) - 1))/(3 - 1) =2( sqrt 3 -1) approx

S= 1/2 * ACAB * sin30^ approx 1 2 *1,46*: