Two corners of a triangle have angles of $(3 pi ) / 4$ and $pi / 12$ . If one side of the triangle has a length of $5$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-05T14:02:38

Largest possible perimeter 28.3196

Sum of the angles of a triangle $=pi$

Two angles are $(3pi)/4, pi/12$
Hence $3^(rd)$ angle is $pi - ((3pi)/4 + pi/12) = pi/6$

We know $a/sin a = b/sin b = c/sin c$

To get the longest perimeter, length 2 must be opposite to angle $pi/12$

$:. 5/ sin(pi/12) = b/ sin((3pi)/4 = c / sin (pi/6)$

$b = (5 sin((3pi)/4))/sin (pi/12) = 13.6603$

$c =( 5* sin(pi/6))/ sin (pi/12) = 9.6593$

Hence perimeter $= a + b + c = 5 + 13.6603 + 9.6593= 28.3196$

Two corners of a triangle have angles of (3 pi ) / 4 (3 pi ) / 4 and pi / 12 pi / 12 . If one side of the triangle has a length of 5 5 , what is the longest possible perimeter of the triangle?