Two corners of a triangle have angles of $(3 pi ) / 8$ and $pi / 4$ . If one side of the triangle has a length of $7$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-20T10:40:02

Longest possible perimeter P = 25.2918

Given $: /_ A = pi /4, /_B = (3pi)/8$

$/_C = (pi - pi /4 - (3pi)/8 ) = (3pi)/8$

To get the longest perimeter, we should consider the side corresponding to the angle that is the smallest.

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

$7 / sin (pi/4) = b / sin ((3pi)/8) = c / sin ((3pi)/8)$

It’s an isosceles triangle as $/_B = /_C = ((3pi)/8)$

$:. b = c = (7 * sin ((3pi)/8)) / sin (pi/4) = 9.1459$

Longest possible perimeter $P = 7 + 9.1459 + 9.1459 = 25.2918$

Two corners of a triangle have angles of (3 pi ) / 8 (3 pi ) / 8 and pi / 4 pi / 4 . If one side of the triangle has a length of 7 7 , what is the longest possible perimeter of the triangle?