A triangle has two corners with angles of $pi / 4$ and $(3 pi )/ 8$ . If one side of the triangle has a length of $2$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-08T12:28:07

Largest possible area of the triangle is 2.4142

Given are the two angles $(3pi)/8$ and $pi/4$ and the length 1

The remaining angle:

$= pi - (((3pi)/8) + pi/4) = (3pi)/8$

I am assuming that length AB (2) is opposite the smallest angle.

Using the ASA

Area $=(c^2*sin(A)*sin(B))/(2*sin(C)$

Area $=( 2^2*sin((3pi)/8)*sin((3pi)/8))/(2*sin(pi/4))$

Area $=2.4142$

A triangle has two corners with angles of pi / 4 pi / 4 and (3 pi )/ 8 (3 pi )/ 8 . If one side of the triangle has a length of 2 2 , what is the largest possible area of the triangle?