Question

在三角形ABC中,内角A,B,C的对边分别为a,b,c,已知b的平方=ac,且cosB=3/4,1、求1/ta如题

Answer 1

1.a,b,c成等比数列,ac=b^2,sinA*sinC=sinB^2 (a/sinA=Bb/sinB=c/sinC=2R) cotA+cotC= cosA /sinA +cosC /sinC =(cosA sinC +cosC sinA )/sinA sinC =sin(A+C )/sinB ^2 =sinB /sin B^2 =1/sinB =根号（1-cosB^2)=根号7/4 2.a、b、c成等比数列,b^2=ac （向量BA）*（向量BC）=|BA|*|BC|cosB=ac*0.75=1.5,ac=2 由余弦定理：b^2=a^2+c^2-2accosB ac=a^2+c^2-1.5ac a^2+c^2=2.5ac=5 (a+c)^2=a^2+c^2+2ac=9 故a+c=3