A triangle has two corners with angles of $\frac{π}{6}$ $pi / 6$ and $\frac{5 π}{8}$ $(5 pi )/ 8$ . 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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A triangle has two corners with angles of $\frac{π}{6}$ $pi / 6$ and $\frac{5 π}{8}$ $(5 pi )/ 8$ . 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 · 2017-01-18T09:04:26

$0.5624 u n i t^{2}$ $0.5624unit^2$

Explanation:

For the largest triangle, the shortest length is reflect to smallest angle.

The angles of triangle are $\frac{π}{6} = \frac{4 π}{24}, \frac{5 π}{8} = \frac{15 π}{24}, and \frac{5 π}{24}$ $pi/6= (4pi)/24, (5pi)/8=(15pi)/24, and (5pi)/24$ .

Therefor the length side for 1 is $\frac{π}{6}$ $pi/6$

Let another length of triangle are A and B.

Find length of A and B.
$\frac{A}{sin (\frac{5 π}{8})} = \frac{1}{sin (\frac{π}{6})}$ $A/sin((5pi)/8) = 1/sin(pi/6)$
$A = \frac{1}{sin (\frac{π}{6})} \cdot sin (\frac{5 π}{8})$ $A = 1/sin(pi/6)*sin((5pi)/8)$
$A = 1.8478$ $A=1.8478$

$\frac{B}{sin (\frac{5 π}{24})} = \frac{1}{sin (\frac{π}{6})}$ $B/sin((5pi)/24) = 1/sin(pi/6)$
$B = \frac{1}{sin (\frac{π}{6})} \cdot sin (\frac{5 π}{24})$ $B = 1/sin(pi/6)*sin((5pi)/24)$
$B = 1.2175$ $B=1.2175$

Therefor the largest area for triangle = $\frac{1}{2} A B sin (\frac{π}{6})$ $1/2ABsin(pi/6)$
$= \frac{1}{2} (1.8478) (1.2175) sin (\frac{π}{6})$ $=1/2(1.8478)(1.2175)sin(pi/6)$
$= \frac{1}{2} (1.8478) (1.2175) (0.5)$ $=1/2(1.8478)(1.2175)(0.5)$

$0.5624 u n i t^{2}$ $0.5624unit^2$

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A triangle has two corners with angles of $\frac{π}{6}$ $pi / 6$ and $\frac{5 π}{8}$ $(5 pi )/ 8$ . If one side of the triangle has a length of $1$ $1$ , what is the largest possible area of the triangle?

1 Answer

Explanation:

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A triangle has two corners with angles of $\frac{π}{6}$ $pi / 6$ and $\frac{5 π}{8}$ $(5 pi )/ 8$ . If one side of the triangle has a length of $1$ $1$ , what is the largest possible area of the triangle?