Question

Two corners of a triangle have angles of `(3 pi )/ 8` and `( pi ) / 6`. If one side of the triangle has a length of `14`, what is the longest possible perimeter of the triangle?

Answer 1

Longest possible perimeter of the triangle is `67.63`

Explanation:

As the two angles of a triangle are `(3pi)/8` and `pi/6`,

the third angle is `pi-(3pi)/8-pi/6=(24pi-9pi-4pi)/24=(11pi)/24`

As the smallest angle is `pi/6`, the perimeter will be longest, if the given side `14` is opposite it. Let it be `a=14` and other two sides be `b` and `c` opposite angles of `(3pi)/8` and `(11pi)/24`.

Now according to sine formula,

`a/sinA=b/sinB=c/sinC`

i.e. `b/sin((3pi)/8)=c/sin((11pi)/24)=14/sin(pi/6)=14/(1/2)=28` and then

`b=28sin((3pi)/8)=28xx0.9239=25.8692`

and `c=28sin((11pi)/24)=28xx0.9914=27.7592`

and perimeter is `14+25.8692+27.7592=67.6284~~67.63`