Question

Two corners of a triangle have angles of `(3 pi ) / 8` and `pi / 6`. If one side of the triangle has a length of `3`, what is the longest possible perimeter of the triangle?

Answer 1

`14.4919`

Explanation:

We will using the triangle sine law.
Given a triangle with angles `A,B,C` and with sides viewed by those angles respectively `l_A,l_B,l_C`, the sine law states
`l_A/(sin(A))=l_B/(sin(B))=l_C/(sin(C))`
Here `A = 3 pi/8, B = pi/6, C = pi-(a+b)=(11 pi)/24`
The three possibilities are
`3/(sin(A))=l_B/(sin(B))=l_C/(sin(C))->l_B = 1.62359, l_C = 3.2194`
`l_A/(sin(A))=3/(sin(B))=l_C/(sin(C))->l_A = 5.54328, l_C = 5.94867`
`l_A/(sin(A))=l_B/(sin(B))=3/(sin(C))->l_A = 2.79555, l_B = 1.51294`
So the maximum perimeter is given by
`l_A+l_B+l_C=14.4919`

Answer 2

Just a comment clarifying approach: Find the shortest length of side and then assign the length of 3 to that side. The max perimeter will then occur.