在三角形ABC中,∠A=60度,∠B,∠C的平分线相交于点I,那么∠BIC=多少度
\u5728\u25b3ABC\u4e2d\uff0c\u2220ABC\u548c\u2220ACB\u7684\u5e73\u5206\u7ebf\u4ea4\u4e8e\u70b9I\uff0c\u82e5\u2220A=60\u00b0\uff0c\u5219\u2220BIC=______\u2235\u2220A=60\u00b0\uff0c\u2234\u2220ABC+\u2220ACB=120\u00b0\uff0c\u53c8\u2235\u2220ABC\u4e0e\u2220ACB\u7684\u5e73\u5206\u7ebf\u4ea4\u4e8e\u70b9I\uff0c\u2234\u2220IBC+\u2220ICB=12\uff08\u2220ABC+\u2220ACB\uff09=60\u00b0\uff0c\u2234\u2220BIC=180\u00b0-\uff08\u2220IBC+\u2220ICB\uff09=120\u00b0\uff0e\u6545\u7b54\u6848\u4e3a\uff1a120\u00b0\uff0e
120\u5ea6\u3002\u89d2a60\u5ea6\uff0c\u89d2c\u4e0e\u89d2b\u76f8\u52a0\u662f120\u3002\u56e0\u4e3a\u89d2c\u4e0e\u89d2b\u88ab\u5e73\u5206\uff0c\u6240\u4ee5\uff0c\u89d2dbc\u52a0dcb\u7b49\u4e8e60\u5ea6\uff0c\u7528\u4e09\u89d2\u5f62\u5185\u89d2\u548c\uff0c180-60=120\u5ea6\u3002
∵∠A=60°∴∠ABC+∠ACB=120°
∵∠B,∠C的平分线相交于点I
∴∠IBC+∠ICB=½﹙∠ABC+∠ACB)=60°
∴∠BIC=180°-60°=120°
120°
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