Two corners of a triangle have angles of $\frac{3 π}{8}$ $(3 pi )/ 8$ and $\frac{π}{2}$ $( pi ) / 2$ . If one side of the triangle has a length of $16$ $16$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-08T15:14:44

Largest possible area of the triangle is 309.0193

Given are the two angles $\frac{π}{2}$ $(pi)/2$ and $\frac{3 π}{8}$ $(3pi)/8$ and the length 16

The remaining angle:

$= π - (\frac{π}{2}) + \frac{3 π}{8}) = \frac{π}{8}$ $= pi - ((pi)/2) + (3pi)/8) = (pi)/8$

I am assuming that length AB (16) 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 $= \frac{16^{2} \cdot sin (\frac{π}{2}) \cdot sin (\frac{3 π}{8})}{2 \cdot sin (\frac{π}{8})}$ $=( 16^2*sin(pi/2)*sin((3pi)/8))/(2*sin(pi/8))$

Area $= 309.0193$ $=309.0193$

Two corners of a triangle have angles of 3π8 (3 pi )/ 8 and π2 ( pi ) / 2 . If one side of the triangle has a length of 16 16 , what is the longest possible perimeter of the triangle?