12. TA CÓ 1 0. .SIN15S  OA OB AOB 21 0. .SIN15S  OA OC AOC 21 0. ....

5.12. Ta có 1

⁰

. .sin15S  OA OB

AOB

21

0

S  OA OC

AOC

2. .sin 30S  OB OC

BOC

2Mặt khác, S

_AOB

S

_AOC

S

_BOC

nên 1 1 1. .sin15 . .sin15 . .2 sin15 cos152OA OB  2OA OC  2OB OC   Do đó ^{OA OB OC}



^



^{2OB OC. cos15 .}  OB OCSuy ra 2 cos15   hay ^{1 1 2}



^{6 2}



^{6 2}OB OC  OA OB OC a a.4 2.