In order to find the value of the angle, you will need to use the law of cosines. Recall that for any triangle, like the one shown below,

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

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

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

Now, since we want to find the value of A, we will need to use $\displaystyle \cos A=\frac{b^2+c^2-a^2}{2bc}$.

Plug in the given values of the triangle.

$\displaystyle \cos A=\frac{2^2+4^2-5^2}{2(2)(4)}=-0.3125$

$\displaystyle A=108.21$ degrees

Make sure to round to $\displaystyle 2$ places after the decimal.

By the Law of Cosines,

$\displaystyle BC^2 = AB^2 +AC^2 - 2\left (AB\right )\left ( AC\right )\cos\angle A$

or, equivalently,

$\displaystyle BC = \sqrt[]{AB^2 +AC^2 - 2\left ( AB\right )\left ( AC\right )\cos \angle A}$

Substitute:

$\displaystyle BC = \sqrt[]{31^2 + 42^2 - 2\left ( 31\right )\left ( 42\right )\cos \left ( 115^{\circ}\right )}$

$\displaystyle \textup{BC}= \sqrt[]{961 + 1764 - 2604\cdot (-0.4226)} \approx \sqrt{2725 + 1100} \approx \sqrt{3825}\approx61.9$

Solve using law of cosines: $\displaystyle c^2 = a^2 + b^2 - 2ab \cdot \cos C$ where C is the angle between sides a and b.

$\displaystyle 8^2 = 3^2 + 10^2 - 2(3)(10) \cos \theta$

$\displaystyle 64 = 109 - 60 \cos \theta$ subtract 109 from both sides 

$\displaystyle - 45 = -60 \cos \theta$ divide by -60

$\displaystyle 0.75 = \cos \theta$ take the inverse cosine using a calculator

$\displaystyle 41.41 = \theta$

Sometimes with law of cosines we have to worry about a second angle that the calculator won't give us. In this case, that would be $\displaystyle 360 - 41.41=318.59$ but that is too big an angle to be in a triangle.

You are given the length of two sides of a triangle and the angle between them; therefore, you should use the Law of Cosines to find $\displaystyle \angle A$, $\displaystyle \angle B$, or, in this case, the length of $\displaystyle c$.

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

Substitute the given values for $\displaystyle a$, $\displaystyle b$, and $\displaystyle \angle C$:

$\displaystyle c^{2}=(1)^{2}+(1.2)^{2}-2\cdot 1\cdot 1.2\cos 150^{\circ}$

At this point, if you are solving for $\displaystyle c$, take the square root of both sides of the equation.

$\displaystyle \sqrt{c^{2}}=\sqrt{(1)^{2}+(1.2)^{2}-2\cdot 1\cdot 1.2\cos 150^{\circ}}$

$\displaystyle c=\sqrt{(1)^{2}+(1.2)^{2}-2\cdot 1\cdot 1.2\cos 150^{\circ}}$

This question merely asks for the equation, rather than the solution, so you need not simplify any further.

Two of the answer choices are equations derived from the Law of Sines. To use the Law of Sines, you must know at least one side and angle that correspond to one another, which is not the case here.

We are given three sides and our desire is to find an angle, this means we must utilize the Law of Cosines. Since the angle desired is $\displaystyle \angle B$ the equation must be rewritten as such:

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

Substituting the given values:

$\displaystyle 5.1^{2} = 6.2^{2} + 3.5^{2} - 2\left(6.2 \right )\left(3.5 \right )\cos B$

Rearranging:

$\displaystyle \cos B = \frac{5.1^{2} - 3.5^{2} - 6.2^{2}}{-2\left(3.5 \right )\left(6.2)}$

Solving the right hand side and taking the inverse cosine we obtain:

$\displaystyle B = 55^{\circ}$

This is a straightforward Law of Cosines problem since we are given three sides and desire one of the corresponding angles in the triangle. We write down the Law of Cosines to start:

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

Substituting the given values:

$\displaystyle 12.08^{2} = 9.72^{2} + 13.91^{2} - 2\left(9.72 \right )\left(13.91 \right )\cos C$

Isolating the angle:

$\displaystyle \cos C = \frac{12.08^{2} - 9.72^{2} - 13.91^{2}}{-2\left(9.72 \right )\left(13.91 \right )}$

The final step is to take the inverse cosine of both sides:

$\displaystyle C = \cos^{-1}\left(\frac{12.08^{2} - 9.72^{2} - 13.91^{2}}{-2\left(9.72 \right )\left(13.91 \right )}\right) = 58.31^{\circ} = 58^{\circ}$

The problem gives the lengths of three sides and asks to find an angle. We can use the Law of Cosines to solve for the angle. Because we are solving for $\displaystyle \small \angle C$, we use the equation:

$\displaystyle \small \small \small c^2 = a^2 + b^2 - 2abcos\angle C$

Substituting the values from the problem gives

$\displaystyle \small 13^2 = 5^2 + 12^2 - 2(5)(12)cos\angle C$

Isolating $\displaystyle \small \angle C$ by itself gives

$\displaystyle \small \small \angle C = cos(\frac{13^2 - 5^2 - 12^2}{(-2)(5)(12)}) = 90^o$

We are given the lengths of the three sides to a triangle. Therefore, we can use the Law of Cosines to find the angle being asked for. Since we are looking for $\displaystyle \small \angle B$ we use the equation,

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

Inputting the values we are given,

$\displaystyle \small 3^2 = 4^2 + 5^2 - 2(4)(5)\cos\angle B$

Next we isolate $\displaystyle \small \angle B$ by itself to solve for it

$\displaystyle \small \angle B = \cos^{-1}(\frac{3^2-4^2-5^2}{(-2)(4)(5)})$

$\displaystyle \small \angle B = 37^o$

Because the problem provides all three sides of the triangle, we can use the Law of Cosines to solve this problem. Since we are solving for $\displaystyle \small \angle A$, we use the equation

$\displaystyle \small a^2 = b^2 + c^2 - 2ac\cos\angle A$

Substitute in the given values

$\displaystyle \small 10^2 = 6.2^2 + 4.8^2 - 2(6.2)(4.8)\cos\angle A$

Isolate $\displaystyle \small \angle A$

$\displaystyle \small \angle A =\cos^{-1}( \frac{10^2-6.2^2-4.8^2}{(-2)(6.2)(4.8)})$

$\displaystyle \small \angle A = 130^o$

Because we are given two sides of a triangle and the corresponding angle of the third side, we can use the Law of Cosines to find the length of side $\displaystyle \small c$. To find side $\displaystyle \small c$ we use

$\displaystyle \small c^2 = a^2 + b^2 - 2ab\cos\angle C$

Taking the square root of both sides gives us

$\displaystyle \small c = \sqrt{a^2+b^2-2ab\cos\angle C}$

Substituting in the values from the problem

$\displaystyle \small c = \sqrt{10^2 +6.7^2 - 2(10)(6.7)\cos45^o}$

$\displaystyle \small c = 7.1$