A triangle has two corners with angles of # ( pi ) / 4 # and # ( 7 pi )/ 12 #. If one side of the triangle has a length of #8 #, what is the largest possible area of the triangle?

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Answer 1 · 2017-05-28T18:15:20

#A = 32sqrt2sin((7pi)/12)#

Find the third angle:

#pi-pi/4-(7pi)/12 = pi/6#

We will get the largest triangle, if we make the smallest angle be opposite the given side:

Let #angle A = pi/6# and side #a = 8#

Let #angle B = pi/4#

Let #angle C = (7pi)/12#

Use the Law of Sines to find the length of side b:

#b/sin(B) = a/sin(A)#

#b = a/sin(A)sin(B)#

#b = 8/sin(pi/6)sin(pi/4)#

#b = 8/0.5sqrt(2)/2#

#b = 8sqrt2#

The height is:

#h = bsin(C)#

#h = 8sqrt2sin((7pi)/12)#

The Area is:

#A = 1/2"base"xx"height"#

We let side a be the base:

#A = 1/2(8)(8sqrt2sin((7pi)/12))#

#A = 32sqrt2sin((7pi)/12)#