Question

Two corners of a triangle have angles of `pi / 8` and `pi / 3`. If one side of the triangle has a length of `2`, what is the longest possible perimeter of the triangle?

Answer 1

The maximum perimeter is: `11.708` to 3 decimal places

Explanation:

When ever possible draw a diagram. It helps to clarify what you are dealing with.

Notice that I have labeled the vertices as with capital letters and the sides with small letter version of that for the opposite angle.

If we set the value of 2 to the smallest length then the sum of sides will be the maximum.

Using the Sine Rule

`a/(sin(A)) = b/(sin(B))=c/(sin(C))`

`=> a/(sin(pi/8))=b/(sin(13/24 pi)) = c/(sin(pi/3))`

Ranking these with the smallest sine value on the left

`=> a/(sin(pi/8))=c/(sin(pi/3)) = b/(sin(13/24 pi))`

So side `a` is the shortest.

Set `a=2`

`=> c=(2sin(pi/3))/(sin(pi/8))" "=" "4.526` to 3 decimal places

`=> b=(2sin(13/24 pi))/(sin(pi/8))=5.182` to 3 decimal places

So the maximum perimeter is: `11.708` to 3 decimal places