A triangle has two corners with angles of # pi / 6 # and # (3 pi )/ 8 #. If one side of the triangle has a length of #1 #, what is the largest possible area of the triangle?

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Answer 1 · 2017-12-08T12:50:02

Largest possible area of the triangle is 0.916

Given are the two angles #(3pi)/8# and #pi/6# and the length 1

The remaining angle:

#= pi - (((3pi)/8) + pi/6 )= (11pi)/24#

I am assuming that length AB (1) is opposite the smallest angle.

Using the ASA

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

Area#=( 1^2*sin((11pi)/24)*sin((3pi)/8))/(2*sin(pi/6))#

Area#=0.916#