Two corners of a triangle have angles of $\frac{7 π}{12}$ $(7 pi )/ 12$ and $\frac{3 π}{8}$ $(3 pi ) / 8$ . 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 · 2018-02-11T12:16:44

Longest possible perimeter $= 30.9562$ $= color(green)(30.9562$

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Given Two angles $ˆ A = (\frac{7 π}{4}), ˆ B = (\frac{3 π}{8})$ $hatA = ((7pi)/4), hatB = ((3pi)/8)$

Third $ˆ C = π - (\frac{7 π}{12}) - (\frac{3 π}{8}) = \frac{π}{24}$ $hatC = pi - ((7pi)/12) - ((3pi)/8) = pi/24$

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

To get longest perimeter, length should correspond to the least $ˆ C$ $hatC$

$:. a / sin ((7pi)/24) = b / sin ((3pi)/8) = 2 / sin (pi/24)$

$a = (2 * sin ((7pi)/12)) / sin(pi/24) = 14.8$

$b = (2 * sin ((3pi)/8)) / sin (pi/24) = 14.1562$

Longest perimeter $= a + b + c = 14.8 + 14..1562 + 2 = 30.9562$

Two corners of a triangle have angles of (7 pi )/ 12 7π12 (7 pi )/ 12 and (3 pi ) / 8 3π8 (3 pi ) / 8 . If one side of the triangle has a length of 2 2 2 , what is the longest possible perimeter of the triangle?