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 $17$ $17$ , what is the longest possible perimeter of the triangle?

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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 $17$ $17$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-10-11T21:05:29

Largest possible perimeter of the triangle $=$ $=$ 63.4449

Explanation:

Three angles of the triangles are $\frac{π}{6}, \frac{π}{6}, \frac{2 π}{3}$ $pi/6, pi/6, (2pi)/3$
Side $a = 17$ $a=17$
$\frac{a}{sin a} = \frac{b}{sin b} = \frac{c}{sin c}$ $a/sin a=b/sin b=c/sin c$
$\frac{17}{sin (\frac{π}{6})} = \frac{b}{sin (\frac{π}{6})} = \frac{c}{sin (\frac{2 π}{3})}$ $17/sin (pi/6)=b/sin(pi/6)=c/sin((2pi)/3)$
Side $b = 17, c = \frac{17 \cdot sin (\frac{2 π}{3})}{sin (\frac{π}{6})}$ $b=17, c=(17*sin ((2pi)/3))/sin(pi/6)$
$c = \frac{17 \cdot sin (\frac{π}{3})}{sin (\frac{π}{6})} = \frac{17 \cdot (\frac{\sqrt{3}}{2})}{\frac{1}{2}}$ $c=(17*sin(pi/3))/sin(pi/6)=(17*(sqrt3/2))/(1/2)$
Side $c = 17 \sqrt{3}$ $c=17sqrt3$

$:.$ Perimeter of the triangle $=17+17+17sqrt3=17(2+sqrt3)$
Perimeter $=$ 63.4449

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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 $17$ $17$ , what is the longest possible perimeter of the triangle?

1 Answer

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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 $17$ $17$ , what is the longest possible perimeter of the triangle?