Question

A triangle has two corners with angles of `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

32+16`sqrt2`

Explanation:

To create a triangle with maximum area, we want the side of length `8` to be opposite of the smallest angle. Let us denote a triangle `DeltaABC` with `angleA` having the smallest angle, `frac(pi)(8)`, or `22.5`. Thus, `BC=8`.

From here, we may find the final angle, (`180-45-22.5=112.5`), and apply law of sines to find side `AB`.

`frac(sin(22.5))(8)=frac(sin(112.5))(x)`, where `x` is the length of `AB`.

Now, we may find the area of the triangle through the formula `Area=frac(1)(2)ab sin(C)`, or `frac(1)(2)AB*BC sin(B)=32+16sqrt2`.