Question

在△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.

（1）若△ABC面积S△ABC=√7/4，求a+c的值；（2）求cosA/sinA+cosC/sinC的值

Answer 1

解：
b²=a²+c²-2accosB
ac=a²+c²-3ac/2
a²+c²=5ac/2
sinB=√7/4
S=(ac*sinB)/2=√7/4
ac=2
所以 (a+c)²=a²+c²+2ac=9ac/2=9
a+c=3
(2)
b/sinB=c/sinC=a/sinA
sinC=csinB/b, sinA=asinB/b
sinA*sinC=acsin²B/b²=sin²B

cosA/sinA+cosC/sinC
=(cosAsinC+cosCsinA)/sinAsinC
=sin(A+C)/sinAsinC
=sinB/sinAsinC
=1/sinB
=1/(√7/4)
=4√7/7

Answer 2

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