Two corners of a triangle have angles of # ( pi )/ 3 # and # ( pi ) / 4 #. If one side of the triangle has a length of # 7 #, what is the longest possible perimeter of the triangle?

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Answer 1 · 2016-10-15T23:02:01

The longest possible perimeter is 25.13

Compute remaining angle, C, is:

#C = pi - pi/4 - pi/3#

#C = (12pi)/12 - (3pi)/12 - (4pi)/12#

#C = (5pi)/12#

Let angle #A# equal the smallest angle, #pi/4#, then angle #B = (pi)/3#.

Let 7 be the length of the side opposite angle A (we represent opposite sides with corresponding lowercase letters), #a = 7#.

When you do this, using the Law of Sines,

#a/sin(A) = b/sin(B) = c/sin(C)#

, to compute the lengths of sides b and c will give the sides that are the largest perimeter.

#b = 7sin(pi/3)/sin(pi/4)#

#b ~~ 8.57#

#c = 7sin((5pi)/12)/sin(pi/4)#

#c ~~ 9.56#

#p = 7 + 8.57 + 9.56#

#p = 25.13#