A triangle has two corners with angles of # ( pi ) / 2 # and # ( 5 pi )/ 12 #. If one side of the triangle has a length of #19 #, what is the largest possible area of the triangle?

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Answer 1 · 2018-06-09T12:11:44

Largest possible area of triangle is #673.64# sq.unit.

Angle between Sides # A and B# is # /_c= pi/2=90^0#

Angle between Sides # B and C# is # /_a= (5 pi)/12=75^0 #

Angle between Sides # C and A# is

# /_b= 180-(90+75)=15^0# For largest area of triangle

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

angle , #/_b :. B=19# 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=19 #

#:. A/sin a=B/sin b or A/sin 75= 19/sin 15#

#:. A=19*sin 75/sin 15 or A ~~ 70.91# .

Now we know sides #A=70.91 , B=19# and their included angle

#/_c = 90^0#. Area of the triangle is #A_t=(A*B* sin c)/2# or

#A_t=(70.91*19* sin 90)/2~~ 673.64# sq.unit

Area of the largest possible triangle is #673.64# sq.unit [Ans]