Question

三角形ABC中,内角A,B,C的对边分别为a,b,c,已知b^2=ac,cosB=3/4.（1）求1/tanA +1/t

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