A triangle has two corners with angles of # pi / 4 # and # pi / 6 #. If one side of the triangle has a length of #1 #, what is the largest possible area of the triangle?

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Answer 1 · 2016-10-20T20:44:39

The largest area is #A ~~ 0.68#

Let #/_ A = pi/6# and #/_ B = pi/4#, then #/_C = pi - pi/4 - pi/6 = (7pi)/12#

Note: We assign the given side to be the side opposite the smallest angle, because that give the largest area.

Let a = the side opposite #/_A# = 1
Let b = the side opposite #/_B#
Let c = the side opposite #/_C#

We can use The Law of Sines to write the following equation:

#b/sin(/_B) = (a)/sin(/_A)#

Solve for b:

#b = (a)sin(/_B)/sin(/_A)#

Choose side #a# to be the base, then the height, h, of the triangle is:

#h = bsin(/_C)#

Substituting for b:

#h = ((a)sin(/_B)/sin(/_A))sin(/_C)#

The area, A, with side a as the base is:

#A = 1/2ah#

Substituting for h:

#A = 1/2a((a)sin(/_B)/sin(/_A))sin(/_C)#

#A = (a^2)(sin(/_B)sin(/_C))/(2sin(/_A))#

#A = (1^2)(sin(pi/4)sin((7pi)/12))/(2sin(pi/6))#

#A ~~ 0.68#