A triangle has two corners with angles of # ( pi ) / 3 # and # ( pi )/ 6 #. If one side of the triangle has a length of #5 #, what is the largest possible area of the triangle?

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Answer 1 · 2018-01-19T12:08:33

Area of the largest possible triangle is #21.65# sq.unit.

Angle between Sides # A and B# is # /_c= pi/3=180/3=60^0#

Angle between Sides # B and C# is # /_a= pi/6=180/6=30^0 :.#

Angle between Sides # C and A# is # /_b= 180-(60+30)=90^0#

For largest area of triangle #5# should be smallest size. which

is opposite to the smallest angle , #:. A=5#. 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=C/sinc# or

#5/sin30=C/sin60 :. C= 5* sin60/sin30~~ 8.66(2dp)#

Now we know sides #A=5 , C=8.66# and their included angle

#/_b = 90^0#. Area of the triangle is #A_t=(A*C*sinb)/2#

#:.A_t=(5*8.66*sin90)/2 ~~ 21.65# sq.unit.

Area of the largest possible triangle is #21.65# sq.unit [Ans]