已知等腰△ABC,PB=3,PC的平方=7,AP=1,∠A=90°,則AB=AC,設AB=AC=M
過P點作PD⊥AC,交AC于D點,作PE⊥AB,交AB于E點,則四邊形PDAE為矩形,PE=DA=X,PD=EA=Y,CD=AC-DA=M-X
根據(jù)勾股定理,得
PA^2=DA^2+PD^2,1=X^2+Y^2.(1)
PC^2=PD^2+CD^2,7=Y^2+(M-X)^2.(2)
解上方程組,得
X=DA=(M^2-6)/(2M)
CD=M-X=M-(M^2-6)/(2M)=(M^2+6)/(2M)
Y=EA=√(1-X ^2)
=√{1- [(M^2-6)/(2M)]^2}
=[√(16M^2-M^4-36)]/(2M)
BE=AB-EA=AB-Y=M-[√(16M^2-M^4-36)]/(2M)
在RT△PBE中,根據(jù)勾股定理,得
PB^2=BE^2+PE^2
9={M-[√(16M^2-M^4-36)]/(2M)}^2+[(M^2-6)/(2M)]^2
解上方程(現(xiàn)時的水平解不了一元八次方方程),得M=AB=AC
再根據(jù)余弦定理,得
AC^2=PA^2+PC^2-2PA*PC*cos∠CPA
M^2=1+7-(2√7)*cos∠CPA
cos∠CPA=(8-M^2)/(2√7)
即可求出∠CPA,計算復雜,如有時間,請自己計算.