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 8, what is the area of the triangle?

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Answer 1 · 2018-05-04T12:23:11

Area of the triangle is #11.71# sq.unit.

Angle between Sides # A and B# is # /_c= (3pi)/4=(3*180)/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/sin a = B/sin b=C/sin c ; B=8 :. B/sin b=C/sin c# or

# 8/sin 30=C /sin 135 or C= 8* (sin 135/sin 30) ~~ 11.31 (2 dp) #

Now we know, sides #B=8,C~~ 11.31# and their included angle

#/_a = 15^0#. Area of the triangle is #A_t=(B*C*sin a)/2#

#:.A_t=(8*11.31*sin 15)/2 ~~ 11.71# sq.unit

Area of the triangle is #11.71# sq.unit [Ans]