A triangle has sides A, B, and C. The angle between sides A and B is #(pi)/4# and the angle between sides B and C is #pi/6#. If side B has a length of 2, what is the area of the triangle?

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Answer 1 · 2017-07-08T15:08:05

The area of the triangle is #=0.74u^2#

The angle between side #A# and #C# is

#=pi-(1/4pi+1/6pi)=pi-5/12pi=7/12pi#

Applying the sine rule to the triangle

#A/sin(1/6pi)=B/sin(7/12pi)=2/sin(7/12pi)=2.07#

#A=2.07*sin(1/2pi)=1.04#

The area of the triangle is

#=1/2*A*B*sin(1/4pi)=1/2*1.04*2*sin(1/4pi)=0.74u^2#