A triangle has two corners with angles of # pi / 12 # and # pi / 3 #. If one side of the triangle has a length of #6 #, what is the largest possible area of the triangle?

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Answer 1 · 2017-07-13T12:40:48

The area of the triangle is #=58.2u^2#

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The angles aare

#hat(C)=1/12pi#

#hat(A)=1/3pi#

The third angle of the triangle is

#hat(B)=pi-(1/12pi+1/3pi)#

#=pi-5/12pi#

#=7/12pi#

Let the length of the side #c=6# which is opposite the smallest angle of the triangle.

Then, we apply the sine rule to the triangle

#b/sin(7/12pi)=a/sin(pi/3)=6/sin(1/12pi)=23.18#

#b=23.18*sin(7/12pi)=22.39#

#a=23.18*sin(pi/3)=20.07#

The area of the triangle is

#=1/2ab sin(hat(C))=1/2*20.07*22.39*sin(pi/12)#

#=58.2#