A triangle has two corners with angles of # pi / 6 # and # (5 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 · 2017-01-18T09:04:26

#0.5624unit^2#

For the largest triangle, the shortest length is reflect to smallest angle.

The angles of triangle are #pi/6= (4pi)/24, (5pi)/8=(15pi)/24, and (5pi)/24#.

Therefor the length side for 1 is #pi/6#

Let another length of triangle are A and B.

Find length of A and B.
#A/sin((5pi)/8) = 1/sin(pi/6)#
#A = 1/sin(pi/6)*sin((5pi)/8)#
#A=1.8478#

#B/sin((5pi)/24) = 1/sin(pi/6)#
#B = 1/sin(pi/6)*sin((5pi)/24)#
#B=1.2175#

Therefor the largest area for triangle =#1/2ABsin(pi/6)#
#=1/2(1.8478)(1.2175)sin(pi/6)#
#=1/2(1.8478)(1.2175)(0.5)#

#0.5624unit^2#