A triangle has two corners with angles of $\frac{π}{2}$ $pi / 2$ and $\frac{π}{8}$ $( pi )/ 8$ . 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 $\frac{π}{2}$ $pi / 2$ and $\frac{π}{8}$ $( pi )/ 8$ . 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 · 2017-10-11T14:12:49

Largest area possible $= 1.2071$ $=1.2071$

Explanation:

The three angles are #(pi/2), (pi/8), ((3pi)/8)
It’s a right angle triangle.

$\frac{a}{sin a} = \frac{b}{sin b} = \frac{c}{sin c}$ $a/sin a=b/sin b=c/sin c$
$\frac{1}{sin (\frac{π}{8})} = \frac{b}{sin (\frac{3 π}{8})} = \frac{c}{sin (\frac{π}{2})}$ $1/sin (pi/8)=b/sin ((3pi)/8)=c/sin(pi/2)$
Hypotenuse
$c = \frac{sin (\frac{π}{2})}{sin (\frac{π}{8})} = \frac{1}{sin (\frac{π}{8})}$ $c=sin(pi/2)/sin((pi)/8)=1/sin((pi)/8)$
$c = 2.6131$ $c=2.6131$

$b = \frac{sin (\frac{3 π}{8})}{sin (\frac{π}{8})}$ $b=sin((3pi)/8)/sin(pi/8)$
$b = 2.4142$ $b=2.4142$

Area $= \frac{a \cdot b}{2} = \frac{1 \cdot 2.4142}{2} = 1.2071$ $=(a*b)/2=(1*2.4142)/2=1.2071$

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A triangle has two corners with angles of $\frac{π}{2}$ $pi / 2$ and $\frac{π}{8}$ $( pi )/ 8$ . 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 $\frac{π}{2}$ $pi / 2$ and $\frac{π}{8}$ $( pi )/ 8$ . If one side of the triangle has a length of $1$ $1$ , what is the largest possible area of the triangle?