Two corners of a triangle have angles of $(3 pi )/ 8$ and $( pi ) / 2$ . 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 · 2017-12-20T06:20:24

Longest possible perimeter of the triangle is 42.1914

Given triangle is a right angle triangle as one of the angles is $pi/2$

Three angles are $pi/2, (3pi)/8, pi/8$

To get the longest perimeter, the side of length 7 should correspond to angle $pi8$ (smallest angle).

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

$7/sin (pi/8) = b/sin ((3pi)/8) =c/sin (pi/2)$

$b = (7 * sin ((3pi)/8)) / (sin (pi/8)) = 16.8995$

$c =( 7 * sin (pi/2)) / sin (pi/8) = 18.2919$

Longest possible perimeter $= (a + b + c) = 7 + 16.8995 + 18.2919 = 42.1914$

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 7 7 , what is the longest possible perimeter of the triangle?