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 $2$ $2$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-21T08:36:09

Largest possible area of triangle 9.0741

Given $: ∠ A = \frac{π}{8} ∠ B = \frac{3 π}{8}$ $: /_ A = pi /8 /_B = (3pi)/8$

$∠ C = (π - \frac{π}{8} - \frac{3 π}{8}) = \frac{π}{2}$ $/_C = (pi - pi /8 - (3pi)/8 ) = (pi)/2$

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

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a / sin A = b / sin B = c / sin C$

$\frac{2}{sin (\frac{π}{8})} = \frac{b}{sin (\frac{3 π}{8})} = \frac{c}{sin (\frac{π}{2})}$ $2 / sin (pi/8) = b / sin ((3pi)/8) = c / sin ((pi)/2)$

$:. b = (2 * sin ((3pi)/8)) / sin (pi/8) = 1.8478$

$c = (2 * sin (pi/2)) / sin (pi/8) = 5.2263$

Longest possible perimeter $P = 2 + 1.8478 + 5.2263 = 9.0741$

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 2 22 , what is the longest possible perimeter of the triangle?