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

1 Answer
SCooke
Dec 24, 2017

#22.6#

Explanation:

Just for me.. I prefer to work in degrees. It won't affect the area.
The area will depend on WHICH side is 8. It can be between the angles, or opposite either one.
#135^o# and #22.5^o#, so the third angle is also #22.5^o#

That means it is an isosceles triangle with either two sides of length 8 or a base of length 8. The base must be between the two smaller angles. IF it is the length 8, the height is #h = 4 xx tan (22.5) = 1.66# and the area is then:
#A = 1/2 xx 8 xx 1.66 = 6.64#

If the sides are 8, the height is #h = 8 xx sin (22.5) = 3.06#
That makes #1/2# the base #b = 3.06/(tan(22.5)) = 7.387#

and the area is then:
#A = 7.387 xx 3.06 = 22.6#

These are the only two possible areas for these conditions, so the maximum one is 22.6.

Fat and squat are larger than tall and thin! :D

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