A triangle has sides A, B, and C. The angle between sides A and B is #pi/6# 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 · 2017-08-12T11:56:01

Area of the triangle is #29.65 # sq.unit.

The angle between sides A and B is #/_c = pi/6 =180/6=30^0#

The angle between sides B and C is #/_a = pi/12 =180/12=15^0#

The angle between sides C and A is

#/_b = 180-(30+15)=135^0 , B=18# Applying sine law we get

#A/sina = B/sinb or A= B*sina/sinb= 18 *sin15/sin135 ~~ 6.59#

Now we have side #A = 6.59 , B=18# and their included angle

#/_c = 30^0# . Area of the triangle is #A_t=(A*B*sinc)/2# or

#A_t=(6.59*18*sin30)/2 ~~ 29.65 # sq.unit [Ans]