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

2 Answers | Add Yours

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question