# Determine the perimeter of triangle ABC if AB=6, B=pi/4, C=pi/6.

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Given ABC is a triangle such that:

B = pi/4 = 45 degree.

C = pi/6 = 30 degrees.

Then, the third angle A = 180 - B - C = 105

==> A = 105 degrees.

Now we know that:

AB / sinc = AC / sinB

==> 6 / sin(30) = AC / sin 45

==> 6/ (1/2) = AC / (sqrt2/2)

==> AC = 12* sqrt2/2 = 6sqrt2

**==> AC = 6sqrt2 = 8.49.**

Also:

AC/ sin B = BC / sinA

==> 6/ (1/2) = BC / sin 105)

==> 12 = BC / 0.9659

**==> BC = 11.59**

**==> P = 6 + 11.59 + 8.49 = 26.08.**

**Then the perimeter of the triangle ABC = 26.08 units.**

To determine the perimeter of the triangle ABC, we'll have to determine all the lengths of it's sides.

For the beginning, we'll determine the angle A. Since the sum of the angles of a triangle is 180 degrees = pi radians, we'll get 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)]

We'll factorize by 6:

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