A triangle has sides A, B, and C. The angle between sides A and B is #pi/4# and the angle between sides B and C is #pi/12#. If side B has a length of 13, what is the area of the triangle?

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Answer 1 · 2017-11-12T11:02:41

The area of the triangle is #17.86 # sq.unit.

Angle between Sides # A and B# is # /_c= (pi)/4=180/4=45^0#

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

Angle between Sides # C and A# is # /_b= 180-(45+15)=120^0#

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=13 :. A/sina=B/sinb# or

#A/sin15=13/sin120 or A = 13* sin15/sin120 ~~ 3.89(2dp)# unit.

Now we know #A=3.89 , B=13# and their included angle

#/_c=45^0#. The area of triangle is #A_t=(A*B*sinc)/2#

#:. A_t=(3.89*13*sin45)/2 ~~17.86 (2dp)# sq.unit.

The area of the triangle is #17.86 (2dp)# sq.unit. [Ans]