In order to solve this problem, we need to find the angles of the triangle. Only then will we be able to find which answer choice is NOT an angle. Using the Law of Cosines we are able to find each angle.

$\displaystyle \cos(A)=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{9^{2}+11^{2}-6^{2}}{2\times 9\times 11}$

$\displaystyle cos(A)=0.83$

$\displaystyle A=\cos^{-1}(0.83)=33.9^{\circ}$

$\displaystyle \cos(B)=\frac{c^{2}+a^{2}-b^{2}}{2ac}=\frac{11^{2}+6^{2}-9^{2}}{2\times 6\times 11}$

$\displaystyle cos(B)=0.575$

$\displaystyle B=\cos^{-1}(0.575)=54.9^{\circ}$

To find angle $\displaystyle C$ we use the formula again or we can remember that the angles in a triangle add up to $\displaystyle 180^{\circ}$.

$\displaystyle C^{\circ}=180^{\circ}-33.9^{\circ}-54.9^{\circ}=91.2^{\circ}$ 

The answer choice that isn't an actual angle of the triangle is $\displaystyle 48.3^{\circ}$.

Since we are given all 3 sides, we can use the Law of Cosines in the angle form:

$\displaystyle cos(A) = \frac{-a^2+b^2+c^2}{2bc}$

$\displaystyle cos(B) = \frac{a^2-b^2+c^2}{2ac}$

$\displaystyle cos(C) = \frac{a^2+b^2-c^2}{2ab}$

Let's start by finding angle A:

$\displaystyle cos(A) = \frac{-(12)^2+5^2+9^2}{2\cdot5\cdot9} = \frac{-38}{90} = -0.4\overline{22}$

$\displaystyle A = cos^{-1}(-0.4\overline{22}) = 115^{\circ}$

Now let's solve for B:

$\displaystyle cos(B) = \frac{12^2-(5)^2+9^2}{2\cdot12\cdot9} = \frac{200}{216} = 0.\overline{925}$

$\displaystyle B = cos^{-1}(0.\overline{925}) = 22^{\circ}$

We can solve for C the same way, but since we now have A and B, we can use our knowledge that all interior angles of a triangle must add up to 180 to find C.

$\displaystyle C = 180^{\circ} - 115^{\circ} - 22^{\circ} = 43^{\circ}$

We are only given sides, so we must use the Law of Cosines. The equation for the Law of Cosines is

$\displaystyle a^2=b^2+c^2-2bc\cos{A}$,

where $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$ are the sides of a triangle and the angle $\displaystyle A$ is opposite the side $\displaystyle a$.

We have three known sides and three unknown angles, so we must write the Law three times, where each equation lets us solve for a different angle. 

To solve for angle $\displaystyle \theta$, we write

$\displaystyle 16^2=12^2+14^2-2(12)(14) \cos{\theta}$ and solve for $\displaystyle \theta$ using the inverse cosine function $\displaystyle \arccos$ on a calculator to get

$\displaystyle \theta \approx 75.5$.

Similarly, for angle $\displaystyle \phi$,

$\displaystyle 14^2=12^2+16^2-2(12)(16) \cos{\phi}$

$\displaystyle \phi \approx 57.9$

and for $\displaystyle \beta$,

$\displaystyle 12^2=14^2+16^2-2(14)(16) \cos{\beta}$

and $\displaystyle \beta \approx 46.6$

We need to use the Law of Cosines for side $\displaystyle c$ then solve for $\displaystyle cosC$.

Therefore, 

$\displaystyle \frac{c^2-a^2-b^2}{-2ab}=cosC$.

Plugging in the information provided, we have:

$\displaystyle \frac{50^2-44^2-29^2}{-2(44)(29)}=cosC$.

Then simplify, $\displaystyle 0.1085423197=cosC$.

To solve for $\displaystyle C$, use $\displaystyle cos^{-1}(.1085423197)=C$.

Solve and then round to the appropriate units: $\displaystyle 83.76870649=83.8$. Therefore, $\displaystyle C=83.8^\circ$.

When all three sides are given, Law of Cosines is appropriate. 

Since 10 is the largest side length, it is opposite the largest angle and thus should be the c-value in the equation below.

$\displaystyle \\* c^2 = a^2 + b^2 - 2ab \cos (C) \\ \\* 10^2 = 5^2 + 7^2 - 2 \cdot 5 \cdot 7 \cos \(C) \\\\* C = \arccos \left(\frac {10^2-5^2-7^2}{2 \cdot 5 \cdot 7}\right) \approx 111.8^\circ$

We need to use the Law of Cosines in order to solve this problem

$\displaystyle b^{2}=a^{2}+c^{2}-2ac\cos(B)$

in this case, $\displaystyle a=6, c=10, B=40^{\circ}$ 

In order to arrive at our answer, we plug the numbers into our formula:

$\displaystyle b^{2}=6^{2}+10^{2}-2(6)(10)\cos (40^{\circ})$

$\displaystyle b^{2}=36+100-91.9$

$\displaystyle b\approx6.6$

Note: we use the "approximately" to indicate the answer is around 6.6. It will vary depending on your rounding.

In order to solve this problem, we need to use the following formula

$\displaystyle a^{2}=b^{2}+c^{2}-2bc\cos(A)$

in this case, $\displaystyle b=11, c=7, A=60^{\circ}$

We plug our numbers into our formula and get our answer:

$\displaystyle a^{2}=11^{2}+7^{2}-2(11)(7)\cos (60^{\circ})$

$\displaystyle a^{2}=121+49-77$

$\displaystyle a^{2}=93$

$\displaystyle a\approx 9.6$

Note: we use the "approximately" to indicate that the answer is around 9.6. It will vary depending on your rounding.

Write the formula for the Law of Cosines.

$\displaystyle c^2=a^2+b^2-2abcos(C)$

Substitute the side lengths of the triangle and the included angle to find the third length.

$\displaystyle c^2=10^2+20^2-2(10)(20)cos(25)$

$\displaystyle c^2=100+400-400cos(25)$

$\displaystyle c^2=500-400(.9063)$

$\displaystyle c^2=137.48$

$\displaystyle c=11.725$

Round this to the nearest integer.

$\displaystyle c=12$

Write the formula for the Law of Cosines.

$\displaystyle c^2=a^2+b^2-2abcos(C)$

Substitute the known values and solve for $\displaystyle c$.

$\displaystyle c^2=10^2+20^2-2(10)(20)cos(30)$

$\displaystyle c^2=100+400-400(\frac{\sqrt3}{2})$

$\displaystyle c^2=500-200\sqrt3$

$\displaystyle c=\sqrt{500-200\sqrt3} =12.3931=12.4$

The Law of Cosines for side $\displaystyle a$ is,

 $\displaystyle a^2=b^2+c^2-2bc\cdot cosA$.

Plugging in the information we know, the formula is,

 $\displaystyle a^2=400^2+375^2-2(400)(375)cos(35)=54879.38671$.

Then take the square of both sides: $\displaystyle a=234.2634985$.

Finally, round to the appropriate units: $\displaystyle a=234.3m$.