...一点P,Q,已知OP:PA=1:2,OQ:OB=3:2,AQ与BP交于点R答:1)设OR=xa+yb,向量AR=OR-OA=(x-1)a+yb,AQ=OQ-OA=-a+(3/5)b,AR∥AQ,∴-(x-1)=(5/3)y.(1)同理BR=OR-OB=xa+(y-1)b,BP=OP-OB=(1/3)a-b,BR∥BP,∴3x=-(y-1).y=1-3x,(2)把(2)代入(1)*3,-3x+3=5-15x,12x=2,x=1/6,代入(2),y=1/2.∴OR...
在三角形ABC知sinA+sinC=psinB(p属于R),且ac=1/4b平方,当P=5/4,b...答:由sinA+sinC=psinB及正弦定理,a+c=bp,(1)a+c=5/4,ac=1/4,(a-c)^=(5/4)^-1=9/16,∴a-c=土3/4,∴(a,c)=(1,1/4)或(1/4,1)。(2)B是锐角,<==>a^+c^>b^,ac=(1/4)b^,∴(a+c)^>3b^/2,∴p=(a+c)/b>√6/2,|a-c|<b,∴(a-c)^<b^,(a...