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?

Question

1350 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-26T08:46:00

21.85252

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