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:

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

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

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

\(\displaystyle a\approx 18.7\)