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

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Answer 1 · 2017-07-03T11:51:18

The area of the triangle is $=19.3u^2$

The third angle of the triangle is

$=1/2pi-3/8pi=1/8pi$

To have the largest possible area, the side of length $4$ is opposite
the smallest angle, i.e, $1/8pi$

Let the side opposite the angle $3/8pi$ be $=a$

Applying the sine rule to the triangle,

$a/sin(3/8pi)=4/sin(1/8pi)$

Therefore,

$a=4*sin(3/8pi)/sin(1/8pi)=9.66$

The area of the triangle is

$A=1/2*4*9.66=19.3$

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 4 4 , what is the largest possible area of the triangle?