Question

△ABC中，内角A,B,C的对边分别为a,b,c已知b^2=ac且cosB=3/4求

.（1）1/tanA+1/tanC的值（2）设向量BA*BC=3/2,求a+c的值

Answer 1

(1)b^2=ac sin^2B=sinAsinC cosB=3/4 sinB=根号7/4 1/tanA +1/tanC=cosA/sinA+cosC/sinC=(cosAsinC+sinAcosC)/sinAsinC=sin(C+A)/sinAsinC=sinB/sinAsinC =1/sinB=4/根号7 (2)|a|*|c|cosB=a*c |a|*|c|=2 cosB=(a^2+c^2-b^2)/2ac 3/4=(a^2+c^2-ac)/2ac 3/4=(a^2+c^2-2)/4 a^2+c^2=5 ac=2 (a^2+2ac+c^2)-2ac=5 (a+c)^2=9 a、c为三角形两边，得a+c=3