A triangle has sides A, B, and C. The angle between sides A and B is #(5pi)/12# and the angle between sides B and C is #pi/12#. If side B has a length of 7, what is the area of the triangle?

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Answer 1 · 2018-01-16T11:19:20

Area of the triangle is #** 6.12# sq.unit.**

Angle between Sides # A and B# is # /_c= (5pi)/12=75^0#

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

Angle between Sides # C and A# is # /_b= 180-(75+15)=90^0#

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=7 :. A/sina=B/sinb# or

#A/sin15=7/sin90 :. A = 7* sin15/sin90 ~~ 1.81(2dp)#unit.

Now we know sides #A=1.81 , B=7# and their included angle

#/_c = 75^0#. Area of the triangle is #A_t=(A*B*sinc)/2#

#:.A_t=(1.81*7*sin75)/2 ~~ 6.12# sq.unit

Area of the triangle is #6.12# sq.unit [Ans]