Two corners of a triangle have angles of $\frac{7 π}{12}$ $(7 pi ) / 12$ and $\frac{π}{6}$ $pi / 6$ . If one side of the triangle has a length of $2$ $2$ , what is the longest possible perimeter of the triangle?

Question

1140 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-20T11:24:21

Longest possible perimeter P = 8.6921

Given $: ∠ A = \frac{π}{6}, ∠ B = \frac{7 π}{12}$ $: /_ A = pi /6, /_B = (7pi)/12$

$∠ C = (π - \frac{π}{6} - \frac{7 π}{12}) = \frac{π}{4}$ $/_C = (pi - pi /6 - (7pi)/12 ) = (pi)/4$

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

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

$\frac{2}{sin (\frac{π}{6})} = \frac{b}{sin (\frac{7 π}{12})} = \frac{c}{sin (\frac{π}{4})}$ $2 / sin (pi/6) = b / sin ((7pi)/12) = c / sin ((pi)/4)$

$:. b = (2 * sin ((7pi)/12)) / sin (pi/6) = 3.8637$

$c = (2 * sin (pi/4)) / sin (pi/6) = 2.8284$

Longest possible perimeter $P = 2 + 3.8637 + 2.8284 = 8.6921$

Two corners of a triangle have angles of (7 pi ) / 12 7π12(7 pi ) / 12 and pi / 6 π6 pi / 6 . If one side of the triangle has a length of 2 22 , what is the longest possible perimeter of the triangle?