郭敦顒回答:在△abc中已知AB=AC=5,BC=6,点M是AC边上靠近A点的一个三等分点,试问在线段BM上是否存在点P,使PC⊥BM,作AD⊥BC于D,则BD=CD=6/2=3,AD=4,sin B=BD/AB=3/6,∴∠B=∠C=53.13°,CM=5×2/3=10/3,在△BCM中,按余弦定理:cosC=(6²+ 100/9-BM²)/(2×6×10/3)=3/5424/9-BM²=24.9BM²=208,BM=4.8074,按正弦定理有:CM/sin∠CBM=BM/sinC=4.8074/0.8=6.00925,CM/sin∠CBM=(10/3)/ sin∠CBM =6 .00925,sin∠CBM=(10/3)/6 .00925=0 .5547,∴∠CBM=33.69°,∠BMC=180°-33.69°-53.13°=93.18°.∴在线段BM上不存在点P,使PC⊥BM.在BM的延长线上存在点P,使PC⊥BP. A P M 93.18
33.69° 53.13°B D C