A triangle has two corners with angles of $\frac{3 π}{4}$ $(3 pi ) / 4$ and $\frac{π}{6}$ $( pi )/ 6$ . If one side of the triangle has a length of $1$ $1$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-07-17T14:27:17

The area is $= 0.68 u^{2}$ $=0.68u^2$

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The angles are

$ˆ A = \frac{1}{6} π$ $hatA=1/6pi$

$ˆ B = \frac{3}{4} π$ $hatB=3/4pi$

$ˆ C = π - (\frac{3}{4} π + \frac{π}{6}) = π - \frac{11}{12} π = \frac{1}{12} π$ $hatC=pi-(3/4pi+pi/6)=pi-11/12pi=1/12pi$

To have the greatest area, the length $= 1$ $=1$ is opposite the angle $ˆ C$ $hatC$

So, $c = 1$ $c=1$

We apply the sine rule to the triangle

$\frac{a}{sin ˆ A} = \frac{c}{sin ˆ C}$ $a/sin hatA=c/sin hatC$

$\frac{a}{sin (\frac{1}{6} π)} = \frac{1}{sin (\frac{1}{12} π)}$ $a/sin(1/6pi)=1/sin(1/12pi)$

$a = \frac{sin (\frac{1}{6} π)}{sin (\frac{1}{12} π)} = 1.93$ $a=sin(1/6pi)/sin(1/12pi)=1.93$

The area of the triangle is

$= \frac{1}{2} \cdot a \cdot c \cdot sin ˆ B$ $=1/2*a*c*sin hatB$

$= \frac{1}{2} \cdot 1 \cdot 1.93 \cdot sin (\frac{3}{4} π)$ $=1/2*1*1.93*sin(3/4pi)$

$= 0.68 u^{2}$ $=0.68u^2$

A triangle has two corners with angles of 3π4 (3 pi ) / 4 and π6 ( pi )/ 6 . If one side of the triangle has a length of 11 , what is the largest possible area of the triangle?