A triangle #DeltaABC# has sides a, b, and c. The angle between sides a and b is #pi/12# and the angle between sides b and c is #pi/12#. If side b has a length of 12, what is the area of the triangle?

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Answer 1 · 2017-06-11T06:55:54

In #DeltaABC#

#/_C=/_A=pi/12 and " side "b=12#

So #/_B=pi-2*pi/12=(5pi)/6#

Now by rule of sine

#a/sinA=b/sinB#

Now area of the

#DeltaABC=1/2axxbxxsinC#

#=1/2b/sinBxxsinAxxbxxsinC#

#=b^2/2xx(sinAsinC)/sinB#

#=b^2/2xx(sin(pi/12)sin(pi/12))/sin((5pi)/6)#

#=12^2/4xx(2sin^2(pi/12))/sin(pi-pi/6)#

#=36xx(1-cos(pi/6))/sin(pi/6)#

#=36xx(1-sqrt3/2)/(1/2)#

#=36(2-sqrt3)# squnit