A triangle has sides A, B, and C. If the angle between sides A and B is #(pi)/12#, the angle between sides B and C is #(5pi)/12#, and the length of B is 10, what is the area of the triangle?

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Answer 1 · 2017-08-17T12:13:29

The area of the triangle is #12.50 # sq.unit .

The angle between sides #A and B# is # /_c = pi/12=180/12= 15^0#

The angle between sides #B and C# is # /_a =(5 pi)/12=(5*180)/12= 75^0#

The angle between sides #C and A# is # /_b = 180-(75+15)=90^0#

We can find side #A# by aplying sine law # A/sina = B/sinb ; B=10#

#A = B* sina/sinb = 10 * sin 75/sin90 ~~ 9.66 #

Now we have Sides #A(9.66) , B(10) # and their included angle

#/_c = 15^0#. The area of the triangle is # A_t= (A*B*sinc) /2 #

or # A_t= (10*9.66*sin 15)/2 = 12.50 # sq.unit [Ans]