Because $\displaystyle \sin C =\frac{AB}{BC}$, we need to first find the length of BC.

Using the Pythagorean theorem, 

$\displaystyle AB^2+AC^2=BC^2$

$\displaystyle BC=\sqrt {AB^2+AC^2}$

$\displaystyle BC=\sqrt {6^2+2^2}$

$\displaystyle BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$\displaystyle BC=2\sqrt10$.

$\displaystyle \sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

We know that the sine of an angle is:

$\displaystyle \frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$\displaystyle sin(50)=\frac{x}{14}$

This allows us to solve for $\displaystyle x$ easily:

$\displaystyle x=14sin(50)=10.72462220366569$

Rounding, this gives us $\displaystyle 10.72$.

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet.

Using the Pythagorean Theorem, we can solve for it:

$\displaystyle 8^2+10^2=h^2{}$

$\displaystyle h=\sqrt{164}=2\sqrt{41}{}$

So, the sine of this angle is:

$\displaystyle \frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}}{}$

Using the foil method, multiply.  Simplify using the Pythagorean identity sin²Θ + cos²Θ = 1 and the double angle identity sin2Θ = 2sinΘcosΘ.

To solve this equation, it is best to remember the mnemonic SOHCAHTOA which translates to Sin = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent. Looking at the problem statement, we are given the side opposite of the angle we are trying to find as well as the hypotenuse. Therefore, we will be using the SOH part of our mnemonic. Inserting our values, this becomes $\displaystyle SIN(\angle A) = \frac{8}{13}$. Then, we can write $\displaystyle \angle A = SIN^{-1}\left ( \frac{8}{13} \right )$. Solving this, we get $\displaystyle \angle A = 38^{\circ}$

You can draw your scenario using the following right triangle:

Recall that the sine of an angle is equal to the ratio of the opposite side to the hypotenuse of the triangle. You can solve for the angle by using an inverse sine function:

$\displaystyle \small \small \Theta = sin^-^1(\frac{12}{15})=53.13010235415598$ or $\displaystyle 53.13$^{\circ}$$.

Recall that the sine of an angle is equal to the ratio of the opposite side to the hypotenuse of the triangle. You can solve for the angle by using an inverse sine function:

$\displaystyle \Theta = sin^-^1(\frac{30}{75})=23.57817847820183$ or $\displaystyle 23.58$^{\circ}$$.

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle.

$\displaystyle \cos = \frac{adjacent}{hypotenuse}$

$\displaystyle \cos\left ( \theta\right ) = \frac{15}{25} = 0.6$

$\displaystyle \arccos\left ( 0.6\right ) = \theta$

$\displaystyle 53.1^{\circ}= \theta$

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. With this information, we can use the cosine function to find the angle.

$\displaystyle \cos = \frac{adjacent}{hypotenuse}$

$\displaystyle \cos\left ( \theta\right ) = \frac{12}{30} = 0.4$

$\displaystyle \arccos\left ( 0.4\right ) = \theta$

$\displaystyle 66.4^{\circ}= \theta$

With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the adjacent and hypotenuse sides of the triangle with relation to the angle. However, if we plug the given values into the formula for cosine, we get:

$\displaystyle \cos = \frac{adjacent}{hypotenuse}$

$\displaystyle \cos\left ( \theta\right ) = \frac{17}{13} = 1.3$

$\displaystyle \arccos\left ( 1.3\right ) = \theta = \textup{no solution}$

This problem does not have a solution. The sides of a right triangle must be shorter than the hypotenuse. A triangle with a side longer than the hypotenuse cannot exist. Similarly, the domain of the arccos function is $\displaystyle \left [ -1, 1\right ]$. It is not defined at 1.3.