几何三角形
如图,等边三角形ABC内一点P,AP=3,BP=4,CP=5,求∠APB的度数.
![](http://pic.ikafan.com/imgp/L3Byb3h5L2h0dHBzL2ltYWdlMTA5LjM2MGRvYy5jbi9Eb3dubG9hZEltZy8yMDIxLzAxLzIyMjMvMjEzNzM4NTI0XzFfMjAyMTAxMjIxMTMxMzEyODEucG5n.jpg)
题型:解答题难度:中档来源:不详答案(找作业答案--->>上魔方格)如图,以AP为边作等边△APD,连接BD.则∠BAD=60°-∠BAP=∠CAP,在△ADB和△APC中,AD=AP.∠BAD=∠CAP,AB=AC∴△ADB≌△APC(SAS)∴BD=PC=5,又PD=AP=3,BP=4∴BP2+PD2=42+32=25=BD2∴∠BPD=90°∴∠APB=∠APD+∠BPD=60°+90°=150°.
![](http://pic.ikafan.com/imgp/L3Byb3h5L2h0dHBzL2ltYWdlMTA5LjM2MGRvYy5jbi9Eb3dubG9hZEltZy8yMDIxLzAxLzIyMjMvMjEzNzM4NTI0XzJfMjAyMTAxMjIxMTMxMzQ0NTkucG5n.jpg)
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