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?

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?

Aaliyah Anderson · Accepted Answer

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

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"#

Wyatt Anderson · Answer

undefined The longest possible perimeter of the triangle is $ 2 + \sqrt{2} $.