A triangle has sides A,B, and C. If the angle between sides A and B is #(pi)/12#, the angle between sides B and C is #pi/4#, and the length of B is 16, what is the area of the triangle?

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Answer 1 · 2018-04-01T10:58:02

Area of the triangle is #27.05# sq.unit.

Angle between Sides # A and B# is # /_c= pi/12=180/12=15^0#

Angle between Sides # B and C# is # /_a= pi/4=180/4=45^0 :.#

Angle between Sides # C and A# is # /_b= 180-(45+15)=120^0#

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=16 :. A/sina=B/sinb# or

#A/sin45=16/sin120 :. A = 16* sin45/sin120 ~~ 13.06(2dp)#unit

Now we know sides #A~~13.06 , B=16# and their included angle

#/_c = 15^0#. Area of the triangle is #A_t=(A*B*sinc)/2#

#:.A_t=(13.06*16*sin15)/2 ~~ 27.05# sq.unit

Area of the triangle is #27.05# sq.unit [Ans]