Question

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?

Answer 1

`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