Question

Two corners of a triangle have angles of `(2 pi )/ 3` and `( pi ) / 4`. If one side of the triangle has a length of `8`, what is the longest possible perimeter of the triangle?

Answer 1

Longest possible perimeter of triangle is `56.63` unit.

Explanation:

Angle between Sides `A and B` is `/_c= (2pi)/3=120^0`

Angle between Sides `B and C` is `/_a= pi/4=45^0 :.`

Angle between Sides `C and A` is

`/_b= 180-(120+45)=15^0`

For longest perimeter of triangle `8` should be smallest side,

the opposite to the smallest angle , `:. B=8`

The sine rule states if `A, B and C` are the lengths of the sides

and opposite angles are `a, b and c` in a triangle, then:

`A/sina = B/sinb=C/sinc ; B=8 :. B/sinb=C/sinc` or

`8/sin15=C/sin120 or C= 8* (sin120/sin15) ~~ 26.77 (2dp)`

Similarly `A/sina=B/sinb` or

`A/sin45=8/sin15 or A= 8* (sin45/sin15) ~~ 21.86 (2dp)`

Longest possible perimeter of triangle is `P_(max)=A+B+C` or

`P_(max)=26.77+8+ 21.86 ~~ 56.63` unit [Ans]