\(\displaystyle \sin X = \frac{OPPOSITE}{HYPOTENUSE}\)

\(\displaystyle \sin C = \frac{AB}{AC}\)

\(\displaystyle \cos\theta\) is defined as the ratio of the adjacent side to the hypotenuse, or in this case \(\displaystyle \frac{y}{x}\). Solving for y gives the correct expression.

In order to find \(\displaystyle \theta\) we need to utilize the given information in the problem.  We are given the opposite and adjacent sides.  We can then, by definition, find the \(\displaystyle \tan\) of \(\displaystyle \theta\) and its measure in degrees by utilizing the \(\displaystyle \arctan\) function.

\(\displaystyle \tan=\frac{opposite}{adjacent}\)

\(\displaystyle \tan=\frac{10}{8}\)

\(\displaystyle \tan=1.25\)

Now to find the measure of the angle using the \(\displaystyle \arctan\) function.

\(\displaystyle \theta=\arctan1.25\)

\(\displaystyle \rightarrow 51.34^{\circ}\)

If you calculated the angle's measure to be \(\displaystyle 0.90^{\circ}}\) then your calculator was set to radians and needs to be set on degrees.

This problem can be easily solved using trig identities.  We are given the hypotenuse \(\displaystyle (15ft)\) and \(\displaystyle \theta (55^{\circ})\).  We can then calculate side \(\displaystyle o\) using the \(\displaystyle \sin\).

\(\displaystyle \sin=\frac{opposite}{hypotenuse}\)

\(\displaystyle \sin55^{\circ}=\frac{o}{15ft}\)

Rearrange to solve for \(\displaystyle o\).

\(\displaystyle o=\sin55^{\circ}*15ft\)

\(\displaystyle \dpi{100} \rightarrow 12.3ft\)

If you calculated the side to equal \(\displaystyle 8.6ft\) then you utilized the \(\displaystyle \cos\) function rather than the \(\displaystyle \sin\).

\(\displaystyle \tan X= \frac{OPPOSITE}{ADJACENT}\)

\(\displaystyle \tan C=\frac{AB}{CB}\)

\(\displaystyle CB= \frac{AB}{\tan C}\)

For this problem, use the law of sines:

\(\displaystyle \frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}=\frac{\text{side opposite }c}{\sin(c)}\).

In this case, we have values that we can plug in:

\(\displaystyle \frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}\)

\(\displaystyle \frac{Z}{\sin(a)}=\frac{Y}{\sin{(b)}}\)

\(\displaystyle \frac{12}{\sin(30^\circ)}=\frac{20}{\sin{(b)}}\)

\(\displaystyle 24=\frac{20}{\sin(b)}\)

\(\displaystyle \sin(b)=\frac{20}{24}\)

\(\displaystyle b=\sin^{-1}(\frac{20}{24})\)

\(\displaystyle b=56.44^\circ\)

First, observe that this figure is clearly not drawn to scale. Now, we can solve using the law of sines:

\(\displaystyle \frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}=\frac{\text{side opposite }c}{\sin(c)}\).

In this case, we have values that we can plug in:

\(\displaystyle \frac{\text{side opposite }a}{\sin(a)}=\frac{\text{side opposite }b}{\sin(b)}\)

\(\displaystyle \frac{Z}{\sin(a)}=\frac{Y}{\sin{(b)}}\)

\(\displaystyle \frac{30.2}{\sin(18.5^\circ)}=\frac{17.2}{\sin{(b)}}\)

\(\displaystyle 95.18=\frac{17.2}{\sin(b)}\)

\(\displaystyle \sin(b)=\frac{17.2}{95.18}\)

\(\displaystyle b=\sin^{-1}(\frac{17.2}{95.18})\)

\(\displaystyle b=10.41^\circ\)

Since \(\displaystyle c=90^\circ\), we know we are working with a right triangle.

That means that \(\displaystyle \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\).

In this problem, that would be:

\(\displaystyle \sin(a)=\frac{\text{Z}}{\text{X}}\)

Plug in our given values:

\(\displaystyle \sin(a)=\frac{8}{12}\)

\(\displaystyle \sin(a)=\frac{2}{3}\)

\(\displaystyle a=\sin^{-1}(\frac{2}{3})\)

\(\displaystyle a=41.8^\circ\)

Notice that these sides fit the pattern of a 30:60:90 right triangle: \(\displaystyle x:x\sqrt{3}:2x\).

In this case, \(\displaystyle x=6\).

Since angle \(\displaystyle a\) is opposite \(\displaystyle x\), it must be \(\displaystyle 30^\circ\).

The pattern for \(\displaystyle 30^\circ:60^\circ:90^\circ\) is that the sides will be \(\displaystyle x:x\sqrt{3}:2x\).

If the side opposite \(\displaystyle 30^\circ\) is \(\displaystyle 7\), then the side opposite \(\displaystyle 60^\circ\) will be \(\displaystyle 7\sqrt{3}\).