Two corners of a triangle have angles of #(5 pi ) / 12 # and # ( pi ) / 8 #. If one side of the triangle has a length of #4 #, what is the longest possible perimeter of the triangle?

2 Answers

#24.459#

Explanation:

Let in #\Delta ABC#, #\angle A={5\pi}/12#, #\angle B=\pi/8# hence

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

#=\pi-{5\pi}/12-\pi/8#

#={11\pi}/24#

For maximum perimeter of triangle , we must consider the given side of length #4# is smallest i.e. side #b=4# is opposite to the smallest angle #\angle B={\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 ({5\pi}/12)}=\frac{4}{\sin (\pi/8)}=\frac{c}{\sin ({11\pi}/24)}#

#a=\frac{4\sin ({5\pi}/12)}{\sin (\pi/8)}#

#a=10.096# &

#c=\frac{4\sin ({11\pi}/24)}{\sin (\pi/8)}#

#c=10.363#

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

#a+b+c#

#=10.096+4+10.363#

#=24.459#

Jul 26, 2018

I will let you do the final calculation.

Explanation:

Sometimes a quick sketch helps in the understanding of the problem. That is the case hear. You only need to approximate the two given angles.
Tony B

It is immediately obvious (in this case) that the shortest length is AC.

So if we set this to the given permitted length of 4 then the other two are at their maximum.

The most straight forward relationship to use is the sine rule.

#(AC)/sin(B)=(AB)/sin(C)=(BC)/sin(A)# giving:

#(4)/sin(pi/8)=(AB)/sin((5pi)/12)=(BC)/sin(A)#

We start be determining the angle A

Known: #/_A+/_B+/_C=pi" radians"=180 #

#/_A+pi/8+(5pi)/12=pi" radians"#

#/_A=11/24 pi" radians" -> 82 1/2" degrees"#

This gives:

#color(brown)((4)/sin(pi/8)=(AB)/sin((5pi)/12)=(BC)/sin((11pi)/24))#

Thus #AB=(4sin((5pi)/12)) /sin(pi/8)#

and #BC=(4sin( (11pi)/24))/sin(pi/8) #

Work these out and add then all up including the given length of 4