A triangle has two corners with angles of $(3 pi ) / 4$ and $( pi )/ 12$ . If one side of the triangle has a length of $11$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-07-21T10:20:27

The area is $=82.6u^2$

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The angles ot the triangle are

$hatA=3/4pi$

$hatB=1/12pi$

$hatC=pi-(3/4pi+1/12pi)=pi-10/12pi=1/6pi$

The side of length $11$ is opposite the smallest angle in the triangle

The smallest angle is $=hatB$

So,

$b=11$

We apply the sine rule to the triangle

$a/sin hatA=b/sin hatB$

$a/sin(3/4pi)=11/sin(1/12pi)$

$a=11*sin(3/4pi)/sin(1/12pi)=30.05$

The area of the triangle is

$area =1/2ab sin hatC=1/2*30.05*11*sin(1/6pi)$

$=82.6u^2$

A triangle has two corners with angles of (3 pi ) / 4 (3 pi ) / 4 and ( pi )/ 12 ( pi )/ 12 . If one side of the triangle has a length of 11 11 , what is the largest possible area of the triangle?