A triangle has two corners with angles of $\frac{π}{12}$ $pi / 12$ and $\frac{π}{3}$ $pi / 3$ . If one side of the triangle has a length of $14$ $14$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-06-30T09:35:10

The area of the triangle is $= 669.4 u^{2}$ $=669.4u^2$

The third angle of the triangle is

$= π - (\frac{1}{12} π + \frac{1}{3} π)$ $=pi-(1/12pi+1/3pi)$

$= π - \frac{5}{12} π$ $=pi-5/12pi$

$= \frac{7}{12} π$ $=7/12pi$

The angles are

$\frac{1}{12} π, \frac{5}{12} π, \frac{7}{12} π$ $1/12pi, 5/12pi, 7/12pi$

To have the greatest area , the side of length $14$ $14$ is opposite the smallest angle

We apply the sine rule

$\frac{14}{sin (\frac{1}{12} π)} = \frac{A}{sin (\frac{5}{12} π)} = \frac{B}{sin (\frac{7}{12} π)}$ $14/sin(1/12pi)=A/sin(5/12pi)=B/sin(7/12pi)$

$A = 14 \frac{sin (\frac{5}{12} π)}{sin (\frac{1}{12} π)} = 52.2$ $A=14sin(5/12pi)/sin(1/12pi)=52.2$

$B = 14 \frac{sin (\frac{7}{12} π)}{sin (\frac{1}{12} π)} = 99.1$ $B=14sin(7/12pi)/sin(1/12pi)=99.1$

The area of the triangle is

$= \frac{1}{2} \cdot A B \cdot sin (\frac{1}{12} π)$ $=1/2*AB*sin(1/12pi)$

$= 0.5 \cdot 52.2 \cdot 99.1 \cdot sin (\frac{1}{12} π)$ $=0.5*52.2*99.1*sin(1/12pi)$

$= 669.4$ $=669.4$

A triangle has two corners with angles of pi / 12 π12 pi / 12 and pi / 3 π3 pi / 3 . If one side of the triangle has a length of 14 1414 , what is the largest possible area of the triangle?