Question

请问：在三角形ABC中，内角A，B，C的对边a，b，c，已知a，b，c成等比数列，cosB＝3／4

在三角形ABC中，内角A，B，C的对边a，b，c，已知a，b，c成等比数列，cosB＝3／4
1.求CotA+cotC的值 2.向量BA乘以向量BC＝3／2，求a＋c的值

Answer 1

解答：
a，b，c成等比数列
b²=ac
b²=a²+c²-2ac*(3/4)
a²+c²=5ac/2
(1) cotA+cotC
=cosA/sinA+cosC/sinC
=(cosAsinC+cosCsinA)/sinAsinC
=sinB/sinAsinCa²+c²=5ac/2a²+c²=5ac/2
=sin²B/(sinAsinBsinC)
=b²/(acsinB)
=1/sinB
=4/√7
=4√7/7
(2)
BA.BC=3/2
ca*cosB=3/2
ac=2
又因为
a²+c²=5ac/2
(a+c)²=9ac/2=9
所以 a+c=3

Answer 2

a，b，c成等比数列
b²=ac
b²=a²+c²-2ac*(3/4)
a²+c²=5ac/2
(1) cotA+cotC
=cosA/sinA+cosC/sinC
=(cosAsinC+cosCsinA)/sinAsinC
=sinB/sinAsinCa²+c²=5ac/2a²+c²=5ac/2
=sin²B/(sinAsinBsinC)
=b²/(acsinB)
=1/sinB
=4/√7
=4√7/7
(2)
BA.BC=3/2
ca*cosB=3/2
ac=2
又因为
a²+c²=5ac/2
(a+c)²=9ac/2=9
所以 a+c=3