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 $9$ $9$ , what is the largest possible area of the triangle?

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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 $9$ $9$ , what is the largest possible area of the triangle?

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Answer 1 · 2018-01-27T14:06:07

$\frac{81}{2 (\sqrt{3} - 1)} or 55.324$ $81/(2(sqrt3-1)) " " or " " 55.324$

Explanation:

First, we need to find the third angle:

$A + B + C = π$ $A + B + C = pi$

$\frac{3 π}{4} + \frac{π}{6} + c = π$ $(3pi)/4 + pi/6 + c = pi$

$\frac{9 π}{12} + \frac{2 π}{12} + c = π$ $(9pi)/12 + (2pi)/12 + c = pi$

$c = \frac{π}{12}$ $c = pi/12$

So our three angles are $\frac{π}{12}$ $pi/12$ , $\frac{π}{6}$ $pi/6$ , and $\frac{3 π}{4}$ $(3pi)/4$ .

To get the largest possible area for the triangle, we want to make $9$ $9$ the smallest side, so that the other two sides can be as big as possible.

And the smallest side is always opposite of the smallest angle. The relation between side and angle is given through the law of sines:

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a/sinA = b/sinB = c/sinC$

If we assume that the side with length $9$ $9$ is opposite the angle $\frac{π}{12}$ $pi/12$ , then we get:

$\frac{a}{sin A} = \frac{9}{sin (\frac{π}{12})} = \frac{9}{\frac{\sqrt{6} - \sqrt{2}}{4}} = \frac{36}{\sqrt{6} - \sqrt{2}}$ $a/sinA = 9/sin(pi/12) = 9/((sqrt6-sqrt2)/4) = 36/(sqrt6-sqrt2)$

Now that we know this constant, we know that all of the other sides and angles must also create this same constant. So, since we have the angles for the other two sides, we can go ahead and solve for one of the other two sides, like this:

$\frac{b}{sin (\frac{3 π}{4})} = \frac{36}{\sqrt{6} - \sqrt{2}}$ $b/sin((3pi)/4) = 36/(sqrt6-sqrt2)$

$\frac{b}{\frac{\sqrt{2}}{2}} = \frac{36}{\sqrt{6} - \sqrt{2}}$ $b/(sqrt2/2) = 36/(sqrt6-sqrt2)$

$b \sqrt{2} = \frac{36}{\sqrt{6} - \sqrt{2}}$ $bsqrt2 = 36/(sqrt6-sqrt2)$

$b = \frac{36}{\sqrt{2} (\sqrt{6} - \sqrt{2})} = \frac{36}{2 (\sqrt{3} - 1)} = \frac{18}{\sqrt{3} - 1}$ $b = 36/(sqrt2(sqrt6-sqrt2)) = 36/(2(sqrt3-1)) = 18/(sqrt3-1)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now, we know two sides, and the angle connecting them. Just for reference, this is what our triangle looks like (not to scale):

Given two sides and the angle between them, we can use this formula to find the area of the triangle:

$A_Delta = 1/2ab sinC$

$A_Delta = 1/2(9)(18/(sqrt3-1))sin(pi/6)$

$A_Delta = 1/2(9)(18/(sqrt3-1))(1/2)$

$A_Delta = 162/(4(sqrt3-1))$

$A_Delta = 81/(2(sqrt3-1)) = 55.324$

Since we made $9$ the smallest side to maximize the other two lengths, this must be the largest possible area for the triangle.

Final Answer

Answer 2 · 2018-01-27T15:33:01

Largest possible area of the triangle $A_t = (1/2) b c sin A$

$color(red)(A_t = 55.3241)$

Explanation:

Third angle $= pi - ((3pi)/4) - (pi/6) = pi / 12$

$a / sin A = b / sin B = c / sin C$

To get the largest area, length 9 should correspond to smallest angle C = $pi/12$

$a / sin ((3pi)/4) = b / sin (pi/6) = 9 / sin (pi/12)$

$a = (9 * sin ((3pi)/4)) / sin (pi/12) = 24.5885$

$b = (9 * sin (pi/6)) / sin (pi/12) = 17.3867$

Largest possible area of the triangle $A_t = (1/2) b c sin A$

$A_t = (1/2) * 17.3867 * 9 * sin ((3pi)/4) = color(red)(55.3241)$

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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 $9$ $9$ , what is the largest possible area of the triangle?

2 Answers

Explanation:

Explanation:

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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 $9$ $9$ , what is the largest possible area of the triangle?