A triangle has two corners with angles of $pi / 2$ and $(3 pi )/ 8$ . If one side of the triangle has a length of $1$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-12-11T06:12:17

Largest possible area of the triangle is 1.2071

Given are the two angles $(3pi)/8$ and $pi/2$ and the length 1

The remaining angle:

$= pi - ((3pi)/8) + pi/2) = (pi)/8$

I am assuming that length AB (1) is opposite the smallest angle.

Using the ASA

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

Area $=( 1^2*sin((3pi)/8)*sin((pi)/2))/(2*sin(pi/8)$

Area $=1.2071$

A triangle has two corners with angles of pi / 2 pi / 2 and (3 pi )/ 8 (3 pi )/ 8 . If one side of the triangle has a length of 1 1 , what is the largest possible area of the triangle?