In a triangle ABC, a,b, and c are the opposite sides of each angles,if Angle B = angle C,and 2b=$${\sqrt[3]{}} = {\sqrt{{\mathtt{3}}}}$$ a,
(1) What is the value of cos A
(2) cos(2A+$${\mathtt{\pi}}$$/4)=?
+130514 I'm assuming that you're saying that 2b = √3 a → a = 2/√3 b
And since angles B and C are equal...the sides opposite are equal.... so, b = c
And by the Law of Cosines
[(2/√3)b]^2 = b^2 + b^2 - 2(b^2)Cos (A)
(4/3)b^2 - 2b^2 -= -2b^2CosA
[2b^2 - (4/3)b^2] / (2 b^2) = Cos A
(2/3)b^2/ (2b^2) = cosA
1/3 = cos A
Cos-1(1/3) = A = about 70.53°
And Cos [2(70.53) + 45] = Cos(186..06) = about -.994
+130514 I'm assuming that you're saying that 2b = √3 a → a = 2/√3 b
And since angles B and C are equal...the sides opposite are equal.... so, b = c
And by the Law of Cosines
[(2/√3)b]^2 = b^2 + b^2 - 2(b^2)Cos (A)
(4/3)b^2 - 2b^2 -= -2b^2CosA
[2b^2 - (4/3)b^2] / (2 b^2) = Cos A
(2/3)b^2/ (2b^2) = cosA
1/3 = cos A
Cos-1(1/3) = A = about 70.53°
And Cos [2(70.53) + 45] = Cos(186..06) = about -.994
