Question

A triangle has two corners with angles of `pi / 4` and `(5 pi )/ 8`. If one side of the triangle has a length of `7`, what is the largest possible area of the triangle?

Answer 1

`A = 41.87`

Explanation:

Given two angles, the third one in a triangle is fixed. In this case it is `pi/8`. The shortest side length will be opposite the smallest angle, which is this one in this case. Therefore, we know that the side of length 7 is opposite the `pi/8` corner.

We now have three angles and a side, and can calculate the other sides using the Law of Sines, and then calculate the height for the area.
https://www.mathsisfun.com/algebra/trig-solving-asa-triangles.html

`a/(sin(2pi/8)) = c/sin C = 7/(sin(pi/8))`
`b/(sin(5pi/8)) = c/sin C = 7/(sin(pi/8))`

`a xx (sin(pi/8)) = 7 xx (sin(2pi/8))`

`b xx (sin(pi/8)) = 7 xx (sin(5pi/8))`

`a xx 0.383 = 7 xx 0.707` ; `a = 12.92`
`b xx 0.383 = 7 xx 0.924` ; `b = 16.89`

We now use Heron's formula for the area:
`A = sqrt(s(s-a)(s-b)(s-c))`

where `s= (a+b+c)/2` or `"perimeter"/2`.
http://mste.illinois.edu/dildine/heron/triarea.html
https://www.mathopenref.com/heronsformula.html

`s= (7 + 12.92 + 16.89)/2 = 18.41`
`A = sqrt(18.41(18.41 - 7)(18.41 - 12.92)(18.41 - 16.89))`

`A = sqrt(18.41(11.41)(5.49)(1.52)`
`A = sqrt(1752.9) = 41.87`