# Triangle ABCgiven AB=6, B=pi/4, C=pi/6, find perimeter of triangle

*print*Print*list*Cite

### 2 Answers

For the triangle ABC AB = 6, B=pi/4, C=pi/6.

As the angles of a triangle have a sum of pi. A = 7*pi/12

Use the property sin A/a = sin B/b = sin C/c

c = 6, C = pi/6, B = pi/4 and A = 7*pi/12

=> sin (pi/6)/6 = sin (pi/4)/b = sin (7*pi/12)/a

=> b = 8.485

=> a = 11.59

**The perimeter of the triangle is 6 + 8.48 + 11.59 = 26.07**

We'll calculate A

A = pi - pi/4 - pi/6

A = (12pi - 3pi - 2pi)/12

A = 7pi/12

A = pi/2 + pi/12

sin pi/12 = sqrt[(1 - cos pi/6)/2]

sin pi/12 = sqrt(2 - sqrt3)/2

We'll apply sine theorem and we'll get:

AB/sin C = AC/sin B

6/sin pi/6 = AC/sin pi/4

AC* 1/2 = 6*sqrt2/2

AC = 6sqrt2

AC/sin B = BC/sin A

6sqrt2/ sin pi/4 = BC/[sqrt(2 - sqrt3)/2]

BC*sqrt2/2 = {6sqrt2*[sqrt(2 - sqrt3)]}/2

BC = 6[sqrt(2 - sqrt3)]

The perimeter P is:

P = AB + AC + BC

P = 6 + 6sqrt2 + 6[sqrt(2 - sqrt3)]

P = 6[1+sqrt2+sqrt(2 - sqrt3)] units