A triangle has sides A,B, and C. If the angle between sides A and B is #(5pi)/12#, the angle between sides B and C is #pi/4#, and the length of B is 14, what is the area of the triangle?

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Answer 1 · 2018-05-18T12:34:38

Area of the triangle is #77.29# sq.unit.

Angle between Sides # A and B# is

# /_c= (5 pi)/12=(5*180)/12=75^0#

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

Angle between Sides # C and A# is # /_b= 180-(75+45)=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=14 :. B/sinb=C/sinc# or

#14/sin 60=C/sin75 or C= 14* (sin 75/sin 60) ~~ 15.61 (2dp) #

Now we know sides #B=14 , C ~~15.61# and their included angle

#/_a = 45^0#. Area of the triangle is #A_t=(B*C*sin a)/2#

#:.A_t=(14*15.61*sin 45)/2 ~~ 77.29 (2 dp)# sq.unit

Area of the triangle is #77.29# sq.unit [Ans]