The Law of Sines is used here since we have Side - Angle - Side. We setup our equation as follows:

$\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B}$

Next, we substitute the known values:

$\displaystyle \frac{10}{\sin 100} = \frac{5}{\sin B}$

Now we cross multiply:

$\displaystyle 10\sin B = 5\sin 100$

Divide by 10 on both sides:

$\displaystyle \sin B = 0.5\sin100$

Finally taking the inverse sine to obtain the desired angle:

$\displaystyle B = sin^{-1}\left(0.5\sin100\right ) = 29.5^{\circ}$

We have two angles and one side, however we do not have $\displaystyle \angle B$. We can determine the angle using the property of angles in a triangle summing to $\displaystyle 180^{\circ}$:

$\displaystyle \angle B = 180^{\circ} - 17^{\circ} - 110^{\circ} = 53^{\circ}$

Now we can simply utilize the Law of Sines:

$\displaystyle \frac{5}{\sin 17} = \frac{b}{\sin 53}$

Cross multiply and divide:

$\displaystyle \frac{5\sin 53}{\sin 17} = b$

Reducing to obtain the final solution:

$\displaystyle b = 13.66$

If we solve for $\displaystyle b$, we can use the Law of Sines to find $\displaystyle Y$.

Since the sum of angles in a triangle equals $\displaystyle 180^{\circ}$,

$\displaystyle a + b + c = 180^{\circ}$

$\displaystyle 35^{\circ} + b + 93^{\circ} = 180^{\circ}$

$\displaystyle b = 52^{\circ}$

Now, using the Law of Sines:

$\displaystyle \frac{\textup{side opposite a}}{\sin\left ( a\right )} = \frac{\textup{side opposite b}}{\sin\left ( b\right )} = \frac{\textup{side opposite c}}{\sin\left ( c\right )}$

$\displaystyle \frac{{X}}{\sin\left ( c\right )} = \frac{{Y}}{\sin\left ( b\right )}$

$\displaystyle \frac{{49}}{\sin\left ( 93^{\circ}\right )} = \frac{{Y}}{\sin\left ( 52^{\circ}\right )}$

$\displaystyle \frac{49}{0.999} \approx \frac{Y}{0.788}$

$\displaystyle Y \approx \frac{49 \cdot 0.788 }{0.999}$

$\displaystyle Y \approx 38.7$

The Law of Sines states

$\displaystyle \frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$

so for a and b, that sets up

$\displaystyle \frac{a}{\sin(35)}=\frac{b}{\sin(60)}, b=a\frac{\sin(60)}{\sin(35)}=1.51a$

To solve, use the law of sines, $\displaystyle \frac{ \sin A }{a } = \frac{ \sin B }{b}$ where a is the side across from the angle A, and b is the side across from the angle B.

$\displaystyle \frac{ \sin \theta }{ 19 } = \frac{ \sin ( 20^o )}{ 7 }$ cross-multiply

$\displaystyle 7 \sin \theta = 19 \cdot \sin (20^o )$ evaluate the right side

$\displaystyle 7 \sin \theta = 6.4984$ divide by 7 

$\displaystyle \sin \theta = 0.9283$ take the inverse sine 

$\displaystyle \theta = \sin ^ {-1 } (0.9283 ) \approx 68.18 ^o$

To solve, use law of sines, $\displaystyle \frac{ \sin A }{a} = \frac{ \sin B }{b}$ where side a is across from angle A, and side b is across from angle B.

In this case, we have a 90-degree angle across from x, but we don't currently know the angle across from the side length 7. We can figure out this angle by subtracting $\displaystyle 25^o$ from $\displaystyle 90^o$:

$\displaystyle 90^o - 25 ^o = 65^o$

Now we can set up and solve using law of sines:

$\displaystyle \frac{ \sin (65^o )}{ 7 } = \frac{ \sin (90 ^o )} { x }$ cross-multiply

$\displaystyle \sin (90 ^o ) \cdot 7 = \sin (65^o ) \cdot x$ evaluate the sines

$\displaystyle 1 \cdot 7 = (0.9063) x$ divide by 0.9063

$\displaystyle 7.72 = x$

The law of sines tells us that $\displaystyle \frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$, where a, b, and c are the sides opposite of angles A, B, and C. In $\displaystyle \triangle DEF$, these ratios can be used to find $\displaystyle \angle F$:

$\displaystyle \frac{sinD}{d}=\frac{sinF}{f}$

$\displaystyle \frac{sin114^\circ}{16}=\frac{sin F}{5}$

$\displaystyle sinF=\frac{5sin114^\circ}{16}$

$\displaystyle F=sin^{-1}\left ( \frac{5sin114^\circ}{16}\right )$

$\displaystyle F\approx16.6^\circ$

The law of sines states that 

$\displaystyle \frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$.

In this triangle, we are looking for the side length c, and we are given angle A, angle B, and side b. The sum of the interior angles of a triangle is $\displaystyle 180^{\circ}$; using subtraction we find that angle C = $\displaystyle 57^{\circ}$.

We can now form a proportion that includes only one unknown, c:

$\displaystyle \frac{sin52^{\circ}}{13}=\frac{sin57^{\circ}}{c}$

Solving for c, we find that 

$\displaystyle c=\frac{13sin57^{\circ}}{sin52^{\circ}}\approx13.84$.

First, find $\displaystyle m\angle C$. The sum of the interior angles of a triangle is $\displaystyle 180^\circ$, so $\displaystyle m\angle C = 180^\circ - 78^\circ - 40^\circ$, or $\displaystyle 62^\circ$.

Using this information, you can set up a proportion to find side b:

$\displaystyle \frac{sinC}{c}=\frac{sinB}{b}$

$\displaystyle \frac{sin62^\circ}{14}=\frac{sin40^\circ}{b}$

$\displaystyle b=\frac{14sin40^\circ}{sin62^\circ}$

$\displaystyle b\approx10.2$

To use the law of sines, first you must find the measure of $\displaystyle \angle C$. Since the sum of the interior angles of a triangle is $\displaystyle 180^\circ$, $\displaystyle m\angle C=84^\circ$.

Law of sines:

$\displaystyle \frac{\sin A}{a}=\frac{\sin C}{c}$

$\displaystyle \frac{\sin 75^\circ}{a}=\frac{\sin 84^\circ}{19.3}$

$\displaystyle a=\frac{19.3 \sin 75^\circ}{\sin 84^\circ}$

$\displaystyle a\approx 18.7$