A triangle has two corners with angles of # pi / 4 # and # (3 pi )/ 8 #. If one side of the triangle has a length of #2 #, what is the largest possible area of the triangle?

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Answer 1 · 2017-12-08T12:28:07

Largest possible area of the triangle is 2.4142

Given are the two angles #(3pi)/8# and #pi/4# and the length 1

The remaining angle:

#= pi - (((3pi)/8) + pi/4) = (3pi)/8#

I am assuming that length AB (2) is opposite the smallest angle.

Using the ASA

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

Area#=( 2^2*sin((3pi)/8)*sin((3pi)/8))/(2*sin(pi/4))#

Area#=2.4142#