Two corners of a triangle have angles of $\frac{5 π}{8}$ $(5 pi )/ 8$ and $\frac{π}{4}$ $( pi ) / 4$ . If one side of the triangle has a length of $4$ $4$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-08T13:33:28

Largest possible area of the triangle is 13.6569

Given are the two angles $\frac{5 π}{8}$ $(5pi)/8$ and $\frac{π}{4}$ $pi/4$ and the length 4

The remaining angle:

$= π - ((\frac{5 π}{8}) + \frac{π}{4}) = \frac{π}{8}$ $= pi - (((5pi)/8) + pi/4) = pi/8$

I am assuming that length AB (4) 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 $= \frac{4^{2} \cdot sin (\frac{π}{4}) \cdot sin (\frac{5 π}{8})}{2 \cdot sin (\frac{π}{8})}$ $=( 4^2*sin(pi/4)*sin((5pi)/8))/(2*sin(pi/8))$

Area $= 13.6569$ $=13.6569$

Two corners of a triangle have angles of 5π8 (5 pi )/ 8 and π4 ( pi ) / 4 . If one side of the triangle has a length of 4 4 , what is the longest possible perimeter of the triangle?