A triangle has two corners with angles of # (2 pi ) / 3 # and # ( pi )/ 4 #. If one side of the triangle has a length of #2 #, what is the largest possible area of the triangle?

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Answer 1 · 2018-02-13T12:12:55

Area of the largest possible triangle is #15.08# sq.unit.

Angle between Sides # A and B# is # /_c= (2pi)/3=120^0#

Angle between Sides # B and C# is # /_a= pi/4=45^0 :.#

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

For largest area of triangle #2# should be smallest side , which

is opposite to the smallest angle #(/_b=15^0)#, i.e #B=2#

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=2 :. A/sina=B/sinb# or

#A/sin45=2/sin15 :. A= 2* sin45/sin15~~ 5.46(2dp)#

Now we know sides #A=5.46 , B=2.0# and their included angle

#/_c = 120^0#. Area of the triangle is #A_t=(A*B*sinc)/2#

#:.A_t=(5.46*cancel2*sin 120)/cancel2 ~~ 4.73# sq.unit.

Area of the largest possible triangle is #15.08# sq.unit [Ans]