1、 Let the vector OA = (3, - root 3), OB = (COS a.sin a), where 0 is less than or equal to a, less than or equal to two-thirds, if the vector AB = root 13, find the value of Tan A / find the maximum area of triangle AOB two, known sequence {an} is the equal ratio sequence of first term A1 = 1, and an ＞ 0, {BN} is the arithmetic sequence of first term 1, and A5 + B3 = 21, A3 + B5 = 13, find the general term formula of {an} {BN}, Finding the first n terms and Sn of sequence {molecule BN denominator 2An}

1、 Let the vector OA = (3, - root 3), OB = (COS a.sin a), where 0 is less than or equal to a, less than or equal to two-thirds, if the vector AB = root 13, find the value of Tan A / find the maximum area of triangle AOB two, known sequence {an} is the equal ratio sequence of first term A1 = 1, and an ＞ 0, {BN} is the arithmetic sequence of first term 1, and A5 + B3 = 21, A3 + B5 = 13, find the general term formula of {an} {BN}, Finding the first n terms and Sn of sequence {molecule BN denominator 2An}

Sqrt (a) stands for A. Ao | stands for distance under the root. Ao stands for vector 1. Vector AB = ob-0a = (COS (a) - 3, sin (a) + sqrt (3)); ab is reduced to: (1) | ab | = sqrt (13 + 2 * sqrt (3) * sin (a) - 6 * cos (a)); 2 * sqrt (3) * sin (a) = 6 * cos (a) Tan (a) = sqrt (3) can be obtained by squaring the above formula