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 34, what is the area of the triangle?

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Answer 1 · 2017-12-24T11:32:18

Area of the triangle is #166.83# 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=34 :. A/sina=B/sinb# or

#A/sin15=34/sin60 :. A = 34* sin15/sin60 ~~ 10.16(2dp)#unit

Now we know sides #A=10.16 , B=34# and their included angle

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

#:.A_t=(10.16*34*sin105)/2 ~~ 166.83# sq.unit

Area of the triangle is #166.83# sq.unit [Ans]