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

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Answer 1 · 2017-10-18T22:47:10

Perimeter $= **38.6455**$

Three angles are $(3pi)/8, pi/6,(11 pi)/24$

Least angle is $pi/6$ and must correspond to side 8 to get the longest possible perimeter.

$8/ sin (pi/6) = b / sin ((3pi)/8 )= c / sin( (11pi)/24)$

$b = (8*sin ((3pi)/8))/sin (pi/6) = 14.7821$

$c = (8* sin ((11pi)/24))/sin (pi/6) = 15.8631$

Perimeter $= 8 + 14.7821 + 15.8631 = 38.6455$

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