A triangle has two corners with angles of # ( pi ) / 4 # and # ( 7 pi )/ 12 #. If one side of the triangle has a length of #12 #, what is the largest possible area of the triangle?

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Answer 1 · 2018-03-02T12:19:32

Area of the largest possible triangle is #98.34# sq.unit.

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

Angle between Sides # B and C# is # /_a= (7pi)/12=105^0 :.#

Angle between Sides # C and A# is

# /_b= 180-(45+105)=30^0#

For largest area of triangle #12# should be smallest side , which

is opposite to the smallest angle #/_b :.B=12#

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 ; A/sina=B/sinb# or

#A/sin105=12/sin30 :. A= (12*sin105)/sin30~~ 23.18# unit

Now we know sides #A=23.18 , B=12# and their included angle

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

#:.A_t=(23.18*12*sin45)/2 ~~98.34# sq.unit.

Area of the largest possible triangle is #98.34# sq.unit [Ans]