Two corners of a triangle have angles of $\frac{7 π}{12}$ $(7 pi ) / 12$ and $\frac{3 π}{8}$ $(3 pi ) / 8$ . If one side of the triangle has a length of $8$ $8$ , 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 $\frac{3 π}{8}$ $(3 pi ) / 8$ . If one side of the triangle has a length of $8$ $8$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-08T13:54:35

Largest possible area of the triangle is 218.7819

Explanation:

Given are the two angles $\frac{7 π}{12}$ $(7pi)/12$ and $\frac{3 π}{8}$ $(3pi)/8$ and the length 8

The remaining angle:

$= π - ((\frac{7 π}{12}) + \frac{3 π}{8}) = \frac{π}{24}$ $= pi - (((7pi)/12) + (3pi)/8) = pi/24$

I am assuming that length AB (8) is opposite the smallest angle.

Using the ASA

Area $= \frac{c^{2} \cdot sin (A) \cdot sin (B)}{2 \cdot sin (C)}$ $=(c^2*sin(A)*sin(B))/(2*sin(C)$

Area $= \frac{8^{2} \cdot sin (\frac{3 π}{8}) \cdot sin (\frac{7 π}{12})}{2 \cdot sin (\frac{π}{24})}$ $=( 8^2*sin((3pi)/8)*sin((7pi)/12))/(2*sin(pi/24))$

Area $= 218.7819$ $=218.7819$

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Two corners of a triangle have angles of $\frac{7 π}{12}$ $(7 pi ) / 12$ and $\frac{3 π}{8}$ $(3 pi ) / 8$ . If one side of the triangle has a length of $8$ $8$ , what is the longest possible perimeter of the triangle?

1 Answer

Explanation:

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Two corners of a triangle have angles of 7π12(7 pi ) / 12 and 3π8 (3 pi ) / 8 . If one side of the triangle has a length of 88 , 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 $\frac{3 π}{8}$ $(3 pi ) / 8$ . If one side of the triangle has a length of $8$ $8$ , what is the longest possible perimeter of the triangle?