AOD in triangle

OD=5, AD=7, angle AOD = 120.

According to sine theorem

AD/sin∠AOD=OD/sin∠OAD

The solution is sin ∠ oad = 5 ∠ 3/ 14, so cos ∠ oad =114.

∠ ODA =60 -∠OAD

sin∠ODA = sin(60-∠OAD)= sin 60 * cos∠OAD-cos 60 * sin∠OAD = 3√3/ 14

In triangle OAD

OA/sin∠ODA=OD/sin∠OAD

The solution is OA=3.

According to S = 1/2bc * Sina.

s△OAD = 1/2 * OD * OA * sin∠AOD = 1/2 * 3 * 5 *√3/2 = 15√3/4

s△OAB = 1/2 * OA * OB * sin∠AOB = 1/2 * 3 * 5 *√3/2 = 15√3/4

S△OBC=S△OAD, S△ OCD =S△OAB.

So s parallelogram = 15√3.