Question

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?

Answer 1

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)`