Two corners of a triangle have angles of $\frac{7 π}{12}$ $(7 pi )/ 12$ and $\frac{π}{8}$ $pi / 8$ . 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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Two corners of a triangle have angles of $\frac{7 π}{12}$ $(7 pi )/ 12$ and $\frac{π}{8}$ $pi / 8$ . 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 · 2017-10-08T23:51:30

Longest possible perimeter $= 11.1915$ $=11.1915$

Explanation:

The three angles are $\frac{7 π}{12}, \frac{π}{8}, \frac{7 π}{24}$ $(7pi)/12,pi/8,(7pi)/24$
Smallest side has length 2 & $∠ \frac{π}{8}$ $/_pi/8$

$\frac{2}{sin (\frac{π}{8})} = \frac{b}{sin (\frac{7 π}{24})} = \frac{c}{sin (\frac{7 π}{12})}$ $2/sin(pi/8)=b/sin((7pi)/24)=c/sin((7pi)/12)$
$b = \frac{2 \cdot sin (\frac{7 π}{24})}{sin (\frac{π}{8})}$ $b=(2*sin((7pi)/24))/sin(pi/8)$
$b = \frac{2 \cdot 0.7934}{0.3827} = 4.1463$ $b=(2*0.7934)/0.3827=4.1463$

$\frac{2}{sin (\frac{π}{8})} = \frac{c}{sin (\frac{7 π}{12})}$ $2/sin(pi/8)=c/sin((7pi)/12)$
$c = \frac{2 \cdot sin (\frac{7 π}{12})}{sin (\frac{π}{8})}$ $c=(2*sin((7pi)/12))/sin(pi/8)$
$c = \frac{2 \cdot 0.9659}{0.3829} = 5.0452$ $c=(2*0.9659)/0.3829=5.0452$

Longest possible perimeter $= 2 + 4.1463 + 5.0452 = 11.1915$ $=2+4.1463+5.0452=11.1915$

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Two corners of a triangle have angles of $\frac{7 π}{12}$ $(7 pi )/ 12$ and $\frac{π}{8}$ $pi / 8$ . 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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Explanation:

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

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Two corners of a triangle have angles of $\frac{7 π}{12}$ $(7 pi )/ 12$ and $\frac{π}{8}$ $pi / 8$ . If one side of the triangle has a length of $2$ $2$ , what is the longest possible perimeter of the triangle?