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:

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

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

$\displaystyle \frac{15}{\sin(30^\circ)}=\frac{Y}{\sin{(45^\circ)}}$

$\displaystyle \frac{15}{\frac{1}{2}}=\frac{Y}{\frac{\sqrt{2}}{2}}$

Cross multiply:

$\displaystyle 15*\frac{\sqrt{2}}{2}=\frac{1}{2}Y$

Multiply both sides by $\displaystyle 2$:

$\displaystyle 15\sqrt{2}=Y$

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:

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

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

$\displaystyle \frac{Z}{\sin(22^\circ)}=\frac{30}{\sin{85^\circ}}$

$\displaystyle \frac{Z}{0.375}=\frac{30}{0.997}$

$\displaystyle \frac{Z}{0.375}=30.09$

$\displaystyle Z=30.09*0.375$

$\displaystyle Z=11.28$

Since we are given $\displaystyle BC$ and want to find $\displaystyle AB$, we apply the Law of Sines, which states, in part,

$\displaystyle \frac{ \sin m \angle A}{B C} = \frac{ \sin m \angle C}{AB}$

$\displaystyle m \angle A = 75^{\circ }$ and

$\displaystyle m \angle C = 180 - \left ( m \angle A + m \angle B \right )= 180 -(75+ 62) = 180 -137 = 43^{\circ }$

Substitute in the above equation:

$\displaystyle \frac{ \sin75^{\circ }}{16} = \frac{ \sin 43^{\circ }}{AB}$

Cross-multiply and solve for $\displaystyle AB$:

$\displaystyle AB\cdot \sin75^{\circ } = 16 \cdot \sin 43^{\circ }$

$\displaystyle AB = \frac{16 \cdot \sin 43^{\circ }}{\sin75^{\circ }} = \frac{16 \cdot \0.6820}{0.9659} \approx 11.3$

Since we are given $\displaystyle m \angle A$ , $\displaystyle AB$, and $\displaystyle BC$, and want to find $\displaystyle m \angle C$, we apply the Law of Sines, which states, in part,

$\displaystyle \frac{ \sin m \angle C}{AB} = \frac{ \sin m \angle A}{B C}$.

Substitute and solve for $\displaystyle \sin m \angle C$:

$\displaystyle \frac{ \sin m \angle C}{16} = \frac{ \sin 66^{\circ }}{23}$

$\displaystyle \frac{ \sin m \angle C}{16} \cdot 16= \frac{ \sin 66^{\circ }}{23}\cdot 16$

$\displaystyle \sin m \angle C = \frac{ \sin 66^{\circ }}{23}\cdot 16 \approx \frac{ 0.9135}{23}\cdot 16\approx 0.6355$

Take the inverse sine of 0.6355:

$\displaystyle m \angle C = \sin^{-1} 0.6355 \approx 39.5^{\circ }$

There are two angles between $\displaystyle 0^{\circ }$ and $\displaystyle 180^{\circ }$ that have any given positive sine other than 1 - we get the other by subtracting the previous result from $\displaystyle 180^{\circ }$:

$\displaystyle 180 - 39.5 = 140.5^{\circ }$

This, however, is impossible, since this would result in the sum of the triangle measures being greater than $\displaystyle 180^{\circ }$. This leaves $\displaystyle 39.5^{\circ }$ as the only possible answer.

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:

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

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

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

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

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

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

$\displaystyle b=10.41^\circ$