A triangle has sides A, B, and C. The angle between sides A and B is #pi/3# and the angle between sides B and C is #pi/12#. If side B has a length of 18, what is the area of the triangle?

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Answer 1 · 2018-07-27T15:58:21

#37.592 \ \text{unit}^2#

Given in #\triangle ABC# ,

#\angle A=\pi/12#

#\angle C=\pi/3#

#\therefore \angle B=\pi-\angle A-\angle C#

#=\pi-\pi/12-\pi/3#

#={7\pi}/12#

Now, using sine rule in #\triangle ABC# as follows

#\frac{A}{\sin\angle A}=\frac{B}{\sin\angle B}#

#\frac{A}{\sin(\pi/12)}=\frac{18}{\sin({7\pi}/12)}#

#A=18\frac{\sin(\pi/12)}{\sin({7\pi}/12)}#

#A=18(2-\sqrt3)#

Now, the area #\Delta# of triangle #\triangle ABC# with sides #A=18(2-\sqrt3)# & #18# & included angle #\angle C=\pi/3# is given as

#\Delta=1/2AB\sin \angle C#

#=1/2(18)(18(2-\sqrt3))\sin(\pi/3)#

#=162(2-\sqrt3)(\sqrt3/2)#

#=81(2\sqrt3-3)#

#=37.592 \ \text{unit}^2#