在三角形ABC中，a=3，b=2*根号6，角B=2角A.1.求cosA的值2.求c边长

Question

在三角形ABC中，a=3，b=2*根号6，角B=2角A.1.求cosA的值 2.求c边长

Answer 1

(1)
B=2A
a/sinA =b/sinB
3/sinA = 2√6/sin2A
=√6/(sinAcosA)
cosA = √6/3
(2)
C= π-A-B
= π - 3A
cosC = -cos3A
= -( 4(cosA)^3 -3cosA)
=-(8√6/9 -√6)
= √6/9
c^2=a^2+b^2-2abcosC
= 9 + 24 - 2(3)(2√6)(√6/9)
= 33 - 8
=25
c=5