A triangle has two corners with angles of pi / 4 and pi / 3 . If one side of the triangle has a length of 8 , what is the largest possible area of the triangle?

1 Answer
Jun 26, 2016

21.85252

Explanation:

Given for triangle ABC, /_B = pi/4, /_C=pi/3.

We know that sum of the 3 angles of a triangle is pi . Hence /_A = pi - pi/3 - pi/4 => /_A = (5pi)/12

As the angles /_B < /_C < /_A , length of side b < side c < side A

To get the largest possible area of the triangle with any one side with 8, we must have the shortest side as 8.

Then b must be 8

b/Sin B = 8/Sin (pi/4) = 8/(sqrt 2/2) = 16/sqrt 2 = 8 sqrt 2

Using the law of Sines,

b/Sin B = a/Sin A = c/Sin C

a = (b/Sin B) * Sin A = 8 sqrt 2 * sin ((5pi)/12) =

c= (b/Sin B) * Sin C = 8 sqrt 2 * sin ((pi)/3)

Area of the triangle = *(a*b*Sin C)/2

= 8 sqrt 2 * sin ((5pi)/12) 8 sin ((pi)/3)

= 21.85252