A triangle has two corners with angles of # pi / 12 # and # (7 pi )/ 8 #. If one side of the triangle has a length of #8 #, what is the largest possible area of the triangle?

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Answer 1 · 2017-12-20T11:32:15

Largest possible area of the triangle #Delta = 24.2822#

Given are the two angles #(7pi)/8# and #pi/12# and the length 8

The remaining angle:

#= pi - (pi/12 + (7pi)/8) = (pi)/24#

I am assuming that length AB (8) is opposite the smallest angle.

Using the ASA

Area#=(c^2*sin(A)*sin(B))/(2*sin(C))#

Area#=( 8^2*sin((pi)/12)*sin((7pi)/8))/(2*sin(pi/24))#

Area#=24.2822#

Answer 2 · 2017-12-20T11:44:15

Area of the largest possible triangle is #24.28# sq.unit.

Angle between Sides # A and B# is # /_c= =15^0#

Angle between Sides # B and C# is # /_a= (7pi)/8=157.5^0 :.#

Angle between Sides # C and A# is

# /_b= 180-(157.5+15)=7.5^0# For largest area of the triangle

#8# should be smallest size , which is opposite to the smallest

angle #/_b= 7.5^0 :. B=8# The sine rule states if #A, B and C# are

the lengths of the sides and opposite angles are #a, b ,c# in a

triangle, then #A/sina = B/sinb=C/sinc ; B=8 :. A/sina=B/sinb# or

#A/sin157.5=8/sin7.5:. A = 8* sin157.5/sin7.5 ~~23.45 (2dp)#unit

Now we know sides #A=23.45 , B=8# and their included angle

#/_c = 15^0#. Area of the triangle is #A_t=(A*B*sinc)/2#

#:.A_t=(23.45*8*sin15)/2 ~~ 24.28# sq.unit

Area of the largest possible triangle is #24.28# sq.unit [Ans]