A triangle has two corners with angles of # pi / 4 # and # pi / 6 #. If one side of the triangle has a length of #3 #, what is the largest possible area of the triangle?

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Answer 1 · 2018-07-15T12:15:05

Largest possible area of triangle is #6.15# sq.unit

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

Angle between Sides # B and C# is # /_a= pi/6=30^0 :.#

Angle between Sides # C and A# is # /_b= 180-(45+30)=105^0#

Length of one side is #3# , For largest area of triangle #3# should

be smallest side , which is opposite to the smallest angle

#/_a=30^0 :. A=3# 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/sin a = B/sin b=C/sin c ; A=3#

# :. A/sin a=B/sin b or 3/sin 30 = B/sin 105# or

#B= 3* sin 105/sin 30 ~~ 5.8# Now we know sides

#A=3 , B=5.8# and their included angle #/_c =45^0#.

Area of the triangle is #A_t=(A*B*sin c )/2#

#:. A_t=(3*5.8*sin 45 )/2 ~~ 6.15# sq.unit

Largest possible area of triangle is #6.15# sq.unit [Ans]