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 $12$ $12$ , 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 $12$ $12$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-11T05:29:02

Largest possible area of the triangle is 144.1742

Explanation:

Given are the two angles $\frac{7 π}{12}$ $(7pi)/12$ and $\frac{π}{8}$ $pi/8$ and the length 1

The remaining angle:

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

I am assuming that length AB (1) 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 $= \frac{12^{2} \cdot sin (\frac{7 π}{24}) \cdot sin (\frac{7 π}{12})}{2 \cdot sin (\frac{π}{8})}$ $=( 12^2*sin((7pi)/24)*sin((7pi)/12))/(2*sin(pi/8))$

Area $= 144.1742$ $=144.1742$

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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 $12$ $12$ , what is the longest possible perimeter of the triangle?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

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Impact of this question

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 $12$ $12$ , what is the longest possible perimeter of the triangle?