A triangle has sides A, B, and C. The angle between sides A and B is #pi/6# and the angle between sides B and C is #pi/12#. If side B has a length of 15, what is the area of the triangle?

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Answer 1 · 2017-11-30T11:34:34

Area of the triangle is #41.18# sq.unit.

Angle between Sides # A and B# is # /_c= pi/6=180/6=30^0#

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

Angle between Sides # C and A# is # /_b= 180-(30+15)=135^0#

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=15 :. A/sina=B/sinb# or

#A/sin15=15/sin135 :. A = 15* sin15/sin135 ~~ 5.49(2dp)#unit

Now we know sides #A=5.49 , B=15# and their included angle

#/_c = 30^0#. Area of the triangle is #A_t=(A*B*sinc)/2#

#:.A_t=(5.49*15*sin30)/2 ~~ 41.18 (2dp)# sq.unit

Area of the triangle is #41.18# sq.unit [Ans]