A triangle has sides A, B, and C. The angle between sides A and B is #(7pi)/12# and the angle between sides B and C is #pi/12#. If side B has a length of 2, what is the area of the triangle?

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Answer 1 · 2017-12-17T14:36:10

Area of the triangle is # 0.58# sq.unit.

Angle between Sides # A and B# is # /_c= (7pi)/12=105^0#

Angle between Sides # B and C# is # /_a= pi/12=180/12=15^0 :.#

Angle between Sides # C and A# is # /_b= 180-(105+15)=60^0#

The sine rule states if #A, B and C# are the lengths of the sides

and opposite angles are #a, b and c# in a triangle, then:

#A/sina = B/sinb=C/sinc ; B=2 :. A/sina=B/sinb# or

#A/sin15=2/sin60 :. A = 2* sin15/sin60 ~~ 0.60(2dp)#unit

Now we know sides #A=0.60 , B=2# and their included angle

#/_c = 105^0#. Area of the triangle is #A_t=(A*B*sinc)/2#

#:.A_t=(0.6*2*sin105)/2 ~~ 0.58# sq.unit [Ans]