A triangle has two corners with angles of # pi / 2 # and # ( 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 · 2017-10-11T14:12:49

Largest area possible #=1.2071#

The three angles are #(pi/2), (pi/8), ((3pi)/8)
It’s a right angle triangle.

#a/sin a=b/sin b=c/sin c#
#1/sin (pi/8)=b/sin ((3pi)/8)=c/sin(pi/2)#
Hypotenuse
#c=sin(pi/2)/sin((pi)/8)=1/sin((pi)/8)#
#c=2.6131#

#b=sin((3pi)/8)/sin(pi/8)#
#b=2.4142#

Area #=(a*b)/2=(1*2.4142)/2=1.2071#