Two corners of a triangle have angles of $\frac{2 π}{3}$ $(2 pi )/ 3$ and $\frac{π}{6}$ $( pi ) / 6$ . If one side of the triangle has a length of $16$ $16$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2018-01-07T03:15:25

Longest possible perimeter of the triangle is $P_{t} = 71.4256$ $color(purple)(P_t = 71.4256)$

Given angles $A = \frac{2 π}{3}, B = \frac{π}{6}$ $A = (2pi)/3, B = pi /6$

$C = π - \frac{2 π}{3} - \frac{π}{6} = \frac{π}{6}$ $C = pi - (2pi)/3 - pi/6 = pi/6$

It’s an isosceles triangle with sides b & c equal.

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To get the longest perimeter, smallest angle (B & C) should correspond to side 16

$\frac{a}{sin (\frac{2 π}{3})} = \frac{16}{sin (\frac{π}{6})}$ $a / sin ((2pi)/3) = 16 / sin (pi/6)$

$a = \frac{16 \cdot sin (\frac{2 π}{3})}{sin (\frac{π}{6})} = 27.7128$ $a = (16 * sin ((2pi)/3)) / sin (pi/6) = 27.7128$

Perimeter $P_{t} = a + b + c = 16 + 27.7128 + 27.7128 = 71.4256$ $P_t = a + b + c = 16 + 27.7128 + 27.7128 = color (purple)(71.4256)$

Longest possible perimeter of the triangle is $P_{t} = 71.4256$ $color(purple)(P_t = 71.4256)$

Two corners of a triangle have angles of 2π3 (2 pi )/ 3 and π6 ( pi ) / 6 . If one side of the triangle has a length of 16 16 , what is the longest possible perimeter of the triangle?