Two corners of a triangle have angles of # (5 pi )/ 8 # and # ( pi ) / 6 #. If one side of the triangle has a length of # 5 #, what is the longest possible perimeter of the triangle?

1 Answer
Harish Chandra Rajpoot
Jul 27, 2018

#20.3264\ \text{unit#

Explanation:

Let in #\Delta ABC#, #\angle A={5\pi}/8#, #\angle B=\pi/6# hence

#\angle C=\pi-\angle A-\angle B#

#=\pi-{5\pi}/8-\pi/6#

#={5\pi}/24#

For maximum perimeter of triangle , we must consider the given side of length #5# is smallest i.e. side #b=5# is opposite to the smallest angle #\angle B={\pi}/6#

Now, using Sine rule in #\Delta ABC# as follows

#\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}#

#\frac{a}{\sin ({5\pi}/8)}=\frac{5}{\sin (\pi/6)}=\frac{c}{\sin ({5\pi}/24)}#

#a=\frac{5\sin ({5\pi}/8)}{\sin (\pi/6)}#

#a=9.2388# &

#c=\frac{5\sin ({5\pi}/24)}{\sin (\pi/6)}#

#c=6.0876#

hence, the maximum possible perimeter of the #\triangle ABC # is given as

#a+b+c#

#=9.2388+5+6.0876#

#=20.3264\ \text{unit#

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