A triangle has two corners with angles of # pi / 4 # and # (5 pi )/ 8 #. If one side of the triangle has a length of #15 #, what is the largest possible area of the triangle?

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Answer 1 · 2018-07-23T12:55:35

Area of the largest possible triangle is #192.05# sq.unit.

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

Angle between sides #BandC# is # /_a= (5pi)/8=900/8=112.5^0 #

Angle between sides #C and A# is

# /_b= 180-(112.5+45)=22.5^0#. For largest area of triangle

#15# should be smallest side , which is opposite to the smallest

angle , i.e #B=15# 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 ; B=15 :. A/sin a=B/sin b#

# :. A= B* sin a/sin b :. A= 15 * sin 112.5/sin 22.5~~ 36.21#

Now we know sides #A=36.21 , B=15# and their included angle

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

#A_t=(36.21*15*sin 45)/2 ~~192.05# sq.unit.

Area of the largest possible triangle is #192.05# sq.unit [Ans]