Question

急需解决三角形外接圆半径问题已知三角形ABC的外接圆的圆心为O,AB=2,AC=3,BC=√7,则向量AO·向量BC等于

Answer 1

cosA=(AB²+AC²-BC²)/(2AB•AC)=1/2

cosB=(AB²+BC²-AC²)/(2AB•BC)= √7/14

故：sinA=√3/2,sinB=3√21/14

又：BC/sinA=2R,故：R=√21/3=∣OA∣=∣OB∣

故：AD=BD=1

故：cos∠BAO=√21/7,sin∠BAO=2√7/7

故：cos∠AEB=-cos(∠B+∠BAO)

= -cosB•cos∠BAO+sinB•sin∠BAO

=5√3/14

故：向量AO•向量BC=∣OA∣•∣BC∣•cos∠AEB=5/2