A triangle has sides A, B, and C. The angle between sides A and B is #(3pi)/4# and the angle between sides B and C is #pi/12#. If side B has a length of 27, what is the area of the triangle?

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Answer 1 · 2018-07-15T19:02:46

#A=1/4*27^2*sqrt(2)*(sqrt(3)+1)#

At first we will compute the third angle:
#pi-3/4*pi-pi/12=(12pi-9pi-pi)/12=pi/6#
with the Theorem of sines weget

#a=(27*sin(pi/6))/(sin(pi/12))#
note that

#sin(pi/6)=1/2#

#sin(pi/12)=(sqrt(3)-1)/(2sqrt(2))#
Now we use the Formula
#A=1/2*a*b*sin(gamma)#

so we get

#A=1/2*27^2*sqrt(2)/(sqrt(3)-1)#
this is

#A=1/4*27^2*(sqrt(3)+1)#

We have used that
#1/(sqrt(3)-1)=(sqrt(3)+1)/2#