Two corners of a triangle have angles of $(3 pi ) / 8$ and $pi / 3$ . If one side of the triangle has a length of $1$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2018-06-02T16:04:46

Longest possible perimeter $color(crimson)(P = 3.25$

$hat A = (3pi)/8, hat B = pi/3, hat C = (7pi)/24$

Least angle #hat C = (7pi)/24 should correspond to the side of length 1 to get the longest possible perimeter.

Applying the law of Sines,

$a / sin A = b / sin B = c / sin C = 1 / sin ((7pi)/24)$

$a = sin ((3pi)/8)* (1 / sin ((7pi)/24) )= 1.16$

$b = sin (pi/3) * (1/sin ((7pi)/24) )= 1.09$

Longest possible perimeter $color(crimson)(P = 1.16 + 1.09 + 1 = 3.25$

Two corners of a triangle have angles of (3 pi ) / 8 (3 pi ) / 8 and pi / 3 pi / 3 . If one side of the triangle has a length of 1 1 , what is the longest possible perimeter of the triangle?