Question

数学题目求解！！！！！

如图,在三角形ABC中,∠A等于60°。BO1,BO2是三角形的三等分线,则∠BO1C等于多少°?∠BO2C等于多少度？

Answer 1

解：
∠BO1C
=180°-(∠O1BC+∠O1CB)
=180°-(2/3∠ABC+2/3∠ACB)
=180°-2/3(∠ABC+∠ACB)
=180°-2/3(180°-∠A)
=180°-2/3*120°
=100°

∠BO2C
=180°-(∠O2BC+∠O2CB)
=180°-(1/3∠ABC+1/3∠ACB)
=180°-1/3(∠ABC+∠ACB)
=180°-1/3(180°-∠A)
=180°-1/3*120°
=140°