Question

Two corners of a triangle have angles of `(5 pi )/ 8` and `( pi ) / 2`. If one side of the triangle has a length of `1`, what is the longest possible perimeter of the triangle?

Answer 1

`"Perimeter "~~6.03" to 2 decimal places"`

Explanation:

Method: assign the length of 1 to the shortest side. Consequently we need to identify the shortest side.

Extend CA to point P

Let `/_ACB=pi/2 ->90^0` Thus triangle ABC is a right triangle.

That being so then `/_CAB+/_ABC = pi/2" thus " /_CAB < pi/2" and "/_ABC < pi/2`

Consequently the other given angle of magnitude `5/8 pi` has to an external angle

Let `/_BAP=5/8 pi=>/_CAB=3/8 pi`

As `/_CAB > /_ABC` then AC < CB
Also as AC < AB and BC < AC, `color(blue)("AC is the shortest length")`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given that AC = 1

Thus for `/_CAB`

`ABcos(3/8 pi)=1`

`color(blue)(AB = 1/cos(3/8 pi)~~2.6131" to 4 decimal places")`
'................................................................................

`color(blue)(tan(3/8 pi)=(BC)/(AC) = (BC)/1=BC~~2.4142" to 4 decimal places")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Perimeter = `1+ 1/cos(3/8 pi)+tan(3/8 pi)`

`~~6.0273" to 4 decimal places"`