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

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Answer 1 · 2017-12-26T15:39:11

Longest possible perimeter of triangle is $56.63$ $56.63$ unit.

Angle between Sides $A and B$ $A and B$ is $∠ c = \frac{2 π}{3} = 120^{0}$ $/_c= (2pi)/3=120^0$

Angle between Sides $B and C$ $B and C$ 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]

Two corners of a triangle have angles of (2 pi )/ 3 2π3 (2 pi )/ 3 and ( pi ) / 4 π4 ( pi ) / 4 . If one side of the triangle has a length of 8 8 8 , what is the longest possible perimeter of the triangle?