Two corners of a triangle have angles of # (3 pi )/ 8 # and # ( pi ) / 2 #. If one side of the triangle has a length of # 9 #, what is the longest possible perimeter of the triangle?

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Answer 1 · 2016-11-17T18:01:33

#Perimeter = 23.5 + 21.7 + 9 = 54.2#

Let #angle C = pi/2#

Let #angle B = (3pi)/8#

Then #angle A = pi - angle C - angle B = pi/8#

Make the length 9 side be "a" so that it is opposite the smallest angle, A; this will give us the longest possible perimeter:

Let side #a = 9#

Find the length of side b, using the Law of Sines:

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

#b = asin(B)/sin(A)#

#b = 9sin(3pi/8)/sin(pi/8) ~~ 21.7 #

Because side c is opposite a right angle, we can use:

#c = sqrt(a^2 + b^2)#

#c = sqrt(9^2 + 21.7^2)#

#c ~~ 23.5#

#Perimeter = 23.5 + 21.7 + 9 = 54.2#