In a triangle ABC, if a = √3 − 1, b = √3 + 1 and C = 60 find the other side and other two angles

Given A = `sqrt(3) - 1`

B = `sqrt(3) + 1`

C = 60°

Using cosine formula 

C2 = a2 + b2 – 2 ab cos C

= `(sqrt(3) - 1)^2 + (sqrt(3) + 1)^2 - 2(sqrt(3) - 1) xx (sqrt(3) + 1)  cos 60^circ`

= `3 -2sqrt(3) + 1 + 3 + 2sqrt(3) + 1 - 2(3 - 1) xx 1/2`

C2 = 8 – 2 = 6

⇒ C = √6

Using since formula

`"a"/sin"A" = "b"/sin"B" = "c"/sin"C"`

`(sqrt(3) - 1)/sin"A" = (sqrt(3) + 1)/sin"B" = sqrt(6)/sin60^circ`

`(sqrt(3) - 1)/sin"A" = sqrt(6)/sin 60^circ`

`(sqrt(3) - 1)/sin"A" = sqrt(6)/(sqrt(3)/2)`

⇒ `(sqrt(3) - 1)/sin"A" = (2sqrt(3) * sqrt(2))/sqrt(3)`

⇒ `(sqrt(3) - 1)/sin"A" = 2sqrt(2)`

⇒ sin A = `(sqrt(3) - 1)/(2sqrt(2))`  ......(1)

sin(45° – 30°) = sin 45° . cos 30° – cos 45° sin 30°

= `1/sqrt(2) * sqrt(3)/2 - 1/sqrt(2) * 1/2`

sin 15° = `(sqrt(3) - 1)/(2sqrt(2))`  ......(2)

From equations (1) and (2), we have

sin A = sin 15°

⇒ A = 15°

In ∆ABC,

We have A + B + C = 180°

15° + B + 60° = 180°

B = 180° – 75°

B = 105°

∴ The required sides and angles are

C = `sqrt(6)`, A = 15°, B = 105°