Two corners of a triangle have angles of $\frac{3 π}{8}$ $(3 pi ) / 8$ and $\frac{π}{8}$ $pi / 8$ . 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 · 2017-12-05T12:11:22

Longest possible perimeter of the $Delta$ is $color(red)(54.2174)$

Three angles are $pi/8, (3pi)/8, pi/2$

$a/ sin A = b / sin B = c / sin C$

To get the largest possible are, smallest angle should correspond to the side of length 8

$8/ sin (pi/8) = b / sin ((pi)/2) = c / sin ((3pi)/8)$

$b = (8*sin (pi/2)) / sin (pi / 8) = (8*1) / (0.3827 ) = 20.9041$

$c = (8* sin ((3pi)/8)) / sin (pi/8) = (8 * (0.9239))/(0.3827) = 19.3133$

Perimeter = a + b + c = 14 + 20.9041 + 19.3133 = color (red)(54.2174)#

Two corners of a triangle have angles of (3 pi ) / 8 3π8(3 pi ) / 8 and pi / 8 π8 pi / 8 . If one side of the triangle has a length of 8 88 , what is the longest possible perimeter of the triangle?