Question

三角形A,B,C中,内角A,B,C的对边分别为a,b,c,已知b²=ac,cosB=3/4

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

Answer 2

1.=6/7^(1/2)
2.=2+2^(1/2)