BO,CO为三角形ABC两外角,角DBC,角BCE的平分线,若角A为X度,则角BOC多少度

角BOC=90°-1/2X°
∵∠DBO=∠CBO,∠BCO=∠ECO
∴∠BCO+∠CBO=1/2(∠DBC+∠ECB)=1/2(180°-∠ABC+180°-∠ACB)=180°-1/2(
∠ABC+∠ACB)
∵角BOC=180°-(∠BCO+∠CBO)=1/2(∠ABC+∠ACB)
又∠ABC+∠ACB=180-X
∴角BOC=1/2(∠ABC+∠ACB)=1/2(180-X)=90°-1/2X°