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 $1$ $1$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2018-02-19T05:58:10

Perimeter of isosceles triangle $P = a + 2 b = 4.464$ $color(green)(P = a + 2b = 4.464$

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$ˆ A = \frac{2 π}{3}, ˆ B = \frac{π}{6}, s i d e = 1$ $hatA = (2pi)/3, hatB = pi/6, side = 1$

To find the longest possible perimeter of the triangle.

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

It’s an isosceles triangle with
$ˆ B = ˆ C = \frac{π}{6}$ $hat B = hat C = pi/6$

Least angle $\frac{π}{6}$ $pi/6$ should correspond to the side 1 to get the longest perimeter.

Applying sine law, $\frac{a}{sin A} = \frac{c}{sin C}$ $a / sin A = c / sin C$

$a = \frac{1 \cdot sin (\frac{2 π}{3})}{sin (\frac{π}{6})} = \sqrt{3} = 1.732$ $a = (1 * sin ((2pi)/3)) / sin (pi/6) = sqrt3 = 1.732$

Perimeter of isosceles triangle $P = a + 2 b = 1 + (2 \cdot 1.732) = 4.464$ $color(green)(P = a + 2b = 1 + (2 * 1.732) = 4.464$

Two corners of a triangle have angles of (2 pi )/ 3 2π3 (2 pi )/ 3 and ( pi ) / 6 π6 ( pi ) / 6 . If one side of the triangle has a length of 1 1 1 , what is the longest possible perimeter of the triangle?