Two corners of a triangle have angles of # (3 pi )/ 8 # and # ( pi ) / 2 #. If one side of the triangle has a length of # 4 #, what is the longest possible perimeter of the triangle?

1 Answer

Answer:

#8+4\sqrt2+4\sqrt{4+2\sqrt2}#

Explanation:

Let in #\Delta ABC#, #\angle A={3\pi}/8#, #\angle B=\pi/2# hence

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

#=\pi-{3\pi}/8-\pi/2#

#={\pi}/8#

For maximum perimeter of triangle , we must consider the given side of length #4# is smallest i.e. side #c=4# is opposite to the smallest angle #\angle C=\pi/8#

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

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

#\frac{a}{\sin ({3\pi}/8)}=\frac{b}{\sin (\pi/2)}=\frac{4}{\sin ({\pi}/8)}#

#a=\frac{4\sin ({3\pi}/8)}{\sin (\pi/8)}#

#a=4(\sqrt2+1)# &

#b=\frac{4\sin ({\pi}/2)}{\sin (\pi/8)}#

#b=4\sqrt{4+2\sqrt2}#

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

#a+b+c#

#=4(\sqrt2+1)+4\sqrt{4+2\sqrt2}+4#

#=8+4\sqrt2+4\sqrt{4+2\sqrt2}#