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 `1`, what is the longest possible perimeter of the triangle?

Answer 1

The longest possible perimeter is approximately `4.8307`.

Explanation:

First, we find the one remaining angle, using the fact that a triangle's angles add up to `pi`:

For `triangle ABC`:

Let `angle A =(3pi)/8`
Let `angle B = pi/6`

Then

`angle C = pi - (3pi)/8 - pi/6`

`color(white)(angle C) = pi - (9pi)/24 - (4pi)/24`

`color(white)(angle C) = (11pi)/24`

For any triangle, the shortest side is always opposite the smallest angle. (Same goes for the longest side and largest angle.)

To maximize the perimeter, the one known side length should be the smallest. So, since `angle B` is the smallest (at `pi/6`), we set `b=1`.

Now we can use the sine law to calculate the remaining two sides:

`sin A/a=sinB/b`

`=>a=b times (sinA)/(sinB)`

`color(white)(=>a)=1*(sin((3pi)/8))/(sin(pi/6))`

`color(white)(=>a)~~ 0.9239/0.5" "" "=1.8478`

A similar formula is used to show `c ~~ 1.9829`.

Adding these three values (of `a`, `b`, and `c`) together will yield the longest possible perimeter for a triangle like the one described:

`P=" "a" "+b+" "c`
`color(white)P~~1.8478+1+1.9829`
`color(white)P=4.8307`

(Since this is a geometry question, you may be asked to provide the answer in exact form, with radicals. This is possible, but a bit tedious for the sake of an answer here, which is why I've given my answer as an approximate decimal value.)