Two corners of a triangle have angles of #(3 pi ) / 4 # and # pi / 6 #. If one side of the triangle has a length of #9 #, what is the longest possible perimeter of the triangle?

1 Answer
May 19, 2016

Longest Possible perimeter is #(9 (1 + sqrt[2] +sqrt[3]))/(sqrt[3] - 1)#

Explanation:

With the given two angles we can find the 3rd angle by using the concept that sum of all three angles in a triangle is #180^@ or pi#:

#(3pi)/4 + pi/6 + x = pi#
#x = pi - (3pi)/4 - pi/6#
#x = pi - (11pi)/12#
#x = pi/12#

Hence, the third angle is #pi/12#

Now, let's say

#/_A = (3pi)/4, /_B = pi/6 and /_C = pi/12#

Using Sine Rule we have,

#(Sin /_A)/a = (Sin /_B)/b = (Sin /_C)/c#

where, a, b and c are the length of the sides opposite to #/_A, /_B and /_C# respectively.

Using above set of equations, we have the following:

#a = a, b = (Sin /_B)/(Sin /_A)*a, c = (Sin /_C)/(Sin /_A)*a#

#or a = a, b=(Sin (pi/6))/(Sin ((3pi)/4))*a, c=(Sin (pi/12))/(Sin ((3pi)/4))*a#

#rArr a = a, b=a/(sqrt2), c=(a*(sqrt(3) - 1))/2#

Now, to find the longest possible perimeter of the triangle

#P = a + b + c#

Assuming, #a = 9#, we have
#a = 9, b = 9/sqrt2 and c = (9*(sqrt(3) - 1))/2#
#rArrP = 9 + 9/(sqrt2) +(9*(sqrt(3) - 1))/2#
#or P = (9 (1 + sqrt[2] + sqrt[3]))/2#
#or P~~18.66 #

Assuming, #b = 9#, we have
#a = 9sqrt2, b = 9 and c = (9*(sqrt(3) - 1))/sqrt2#
#rArrP = 9sqrt2 + 9 +(9*(sqrt(3) - 1))/sqrt2#
#or P = (9 (2 + sqrt[2] + sqrt[6]))/2#
#or P~~26.39 #

Assuming, #c = 9#, we have
#a = 18/(sqrt3 - 1), b = (9sqrt2)/(sqrt3 - 1) and c = 9#
#rArrP = 18/(sqrt3 - 1) + (9sqrt2)/(sqrt3 - 1) +9#
#or P = (9 (1 + sqrt[2] +sqrt[3]))/(sqrt[3] - 1)#
#or P~~50.98 #

Therefore, Longest possible perimeter of the given triangle is #(9 (1 + sqrt[2] +sqrt[3]))/(sqrt[3] - 1)#