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

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Answer 1 · 2018-01-29T06:58:35

$P = 4.8284 + 5.2263 + 2 = color(purple)(13.0547)$

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Given $A = (3pi)/8, B = (pi)/2$

$C = pi - (3pi)/8 - pi/2 = pi/8$

To get the longest perimeter, side 2 should correspond to the least angle $pi/8$

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

$a = (2 sin ((3pi)/8)) / sin (pi/8) = 4.8284$

$b = (2 sin (pi/2)) / sin (pi/8) = 5.2263$

Longest Perimeter $P = a + b + c$

$P = 4.8284 + 5.2263 + 2 = color(purple)(13.0547)$

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