A triangle has sides A, B, and C. If the angle between sides A and B is #(7pi)/12#, the angle between sides B and C is #pi/3#, and the length of B is 1, what is the area of the triangle?

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Answer 1 · 2018-01-23T12:06:43

Area of the triangle is #1.62# sq.unit.

Angle between Sides # A and B# is # /_c= (7pi)/12=(7*180)/12=105^0#

Angle between Sides # B and C# is # /_a= pi/3=180/3=60^0 :.#

Angle between Sides # C and A# is # /_b= 180-(105+60)=15^0#

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=1 :. B/sinb=C/sinc# or

#1/sin15=C/sin105 or C= 1* (sin105/sin15) ~~ 3.73 (2dp) #

Now we know sides #B=1 , C=3.73# and their included angle

#/_a = 60^0#. Area of the triangle is #A_t=(B*C*sina)/2#

#:.A_t=(1*3.73*sin60)/2 ~~ 1.62(2dp)# sq.unit.

Area of the triangle is #1.62# sq.unit [Ans]