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)