Question

△ABC中,角A,B,C的对边分别是a,b,c,已知b2=ac,且cosB=3/4；

（1）求1/tanA+1/tanC的值；（2）设向量BA·向量BC=3/2，求a+c的值。

Answer 1

(1)由正弦定理得：a=2RsinA,b=2RsinB,c=2RsinC,而b^2=ac,所以(sinB)^2=sinAsinC.
由cosB=3/4得：sinB=√7/4.sin(A+C)=sin(π-B)=sinB.
1/tanA+1/tanC
=cosA/sinA+cosC/sinC
=(sinCcosA+cosCsinA)(sinAsinC)
=sin(A+C)/(sinAsinC)
=sinB/(sinAsinC)
=1/sinB
=4√7/7
(2)BA向量乘BC向量=accosB=(3/4)ac=(3/4)b^2=3/2,ac=b^2=2.
由余弦定理得：a^2+c^2=b^2+2accosB=23=5.
(a+c)^2=a^2+c^2+2ac=5+4=9,所以a+c=3.