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?

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?

Aurora Ayala · Accepted Answer

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

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

Aurora Ayala · Answer

undefined The longest possible perimeter of the triangle is $2 + 2\sqrt{3} + 4$.