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

1 Answer
Mar 9, 2018

Answer:

#9/4(sqrt{2}+1)#

Explanation:

Two of the angles given are #pi/4# and #(3pi)/8#. The third angle is then #pi - pi/4-(3pi)/8 = (3pi)/8#. Thus the triangle is an isosceles triangle. All triangles satisfying this must be similar.

Now, for the area to be as large as possible, its sides must be as large as possible. For this to happen, the one side that is specified must be the smallest one - and hence must be opposite the smallest angle - in this case #pi/4#. So, the largest triangle fitting the specifications is an isosceles triangle with base 3, and both base angles #(3pi)/8#.

The height of this triangle #= 1.5 tan((3pi)/8) = 3/2(sqrt{2}+1)#

Thus the area is #1/2 times 3 times 3/2(sqrt{2}+1) = 9/4(sqrt{2}+1)#

Note :

#tan theta = sin theta/cos theta = (2sin^2theta)/(2sin theta cos theta) = (1-cos(2theta))/sin(2theta) = csc(2theta)-cot(2theta) implies#
#tan((3pi)/8) = csc((3pi)/4)-cot((3pi)/4) = sqrt(2)-(-1) = sqrt(2)+1#