A triangle has sides A, B, and C. The angle between sides A and B is #(3pi)/4# and the angle between sides B and C is #pi/12#. If side B has a length of 3, what is the area of the triangle?

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Answer 1 · 2017-12-27T11:22:12

Area of the triangle is #1.65# sq.unit.

Angle between Sides # A and B# is # /_c= (3pi)/4==135^0#

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

Angle between Sides # C and A# is # /_b= 180-(135+15)=30^0#

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=3 :. A/sina=B/sinb# or

#A/sin15=3/sin30 or A= 3* (sin15/sin30) ~~ 1.55 (2dp) #

Now we know sides #A~~1.55 , B=3# and their included angle

#/_c = 135^0#. Area of the triangle is #A_t=(A*B*sinc)/2#

#:.A_t=(1.55*3*sin135)/2 ~~ 1.65# sq.unit

Area of the triangle is #1.65# sq.unit [Ans]