The Law of Sines is a set of ratios that allows one to compute missing angles and side lengths of oblique triangles (non right angle triangles).

The Law of Sines: 

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

To find the missing angle A we subtract the sum of the two known angles from 180° as the interior angles of all triangles equal 180°.

\(\displaystyle 180^{\circ}-135^{\circ}=45^{\circ}\)

The LOS can be rearranged to solve for the missing side.

\(\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}\rightarrow a=\frac{b}{\sin B}\cdot\sin A\rightarrow \frac{25}{\sin 30}\cdot\sin 45=35.4ft\)

\(\displaystyle \frac{b}{\sin B}=\frac{c}{\sin C}\rightarrow c=\frac{b}{\sin B}\cdot\sin C\rightarrow c=\frac{25}{\sin 30}\cdot\sin 105=48.3ft\)

This is a straightforward Law of Sines problem as we are given two angles and a corresponding side:

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

Substituting the known values:

\(\displaystyle \frac{a}{\sin 39} = \frac{19}{\sin 100}\)

Solving for the unknown side:

\(\displaystyle a = \sin 39 \left(\frac{19}{\sin 100} \right ) = 12.14 = 12\)

This is a straightforward Law of Sines problem since we are given one angle and two sides and are asked to determine the corresponding angle.

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

Substituting the given values:

\(\displaystyle \frac{100}{\sin A} = \frac{802}{\sin 72^{\circ}}\)

Now rearranging the equation:

\(\displaystyle \sin A = \frac{100\sin 72^{\circ}}{802}\)

The final step is to take the inverse sine of both sides:

\(\displaystyle A = \sin^{-1}\left(\frac{100\sin 72^{\circ}}{802}\right) = 6.81^{\circ} = 7^{\circ}\)

We are given two angles and the length of the corresponding side to one of those angles. Because the problem is asking for the corresponding length of the other angle we can use the Law of Sines to find the length of the side \(\displaystyle \small a\). The equation for the Law of Sines is 

\(\displaystyle \small \frac{a}{\sin\angle A} = \frac{b}{\sin\angle B}\)

If we rearrange the equation to isolate \(\displaystyle a\) we obtain

\(\displaystyle \small a = \frac{b\sin\angle A}{\sin\angle B}\)

Substituting on the values given in the problem

\(\displaystyle \small a = \frac{4\sin40^o}{\sin2^o}\)

\(\displaystyle \small a = 74\)

Because we are given the two angles and the length of the corresponding side to one of those angles, we can use the Law of Sines to find the length of the side that we need. So we use the equation

\(\displaystyle \small \frac{b}{\sin\angle B} = \frac{c}{\sin\angle C}\)

Rearranging the equation to isolate \(\displaystyle \small b\) gives

\(\displaystyle \small b = \frac{c\sin\angle B}{\sin\angle C}\)

Substituting in the values from the problem gives

\(\displaystyle \small b = \frac{18\sin22^o}{\sin7^o}\)

\(\displaystyle \small b = 55\)

We can use the Law of Sines to find the length of the missing side, because we have its corresponding angle and the length and angle of another side. The equation for the Law of Sines is 

\(\displaystyle \small \frac{a}{\sin\angle A} = \frac{b}{\sin\angle B}\)

Isolating \(\displaystyle \small a\) gives us

\(\displaystyle \small a = \frac{b\sin\angle A}{\sin\angle B}\)

Finally, substituting in the values of the of from the problem gives

\(\displaystyle \small a= \frac{3.5\sin40^o}{\sin35^o}\)

\(\displaystyle \small a = 4\)

To solve, use Law of Sines, \(\displaystyle \frac{a}{sinA} = \frac{b}{sinB}\), where A is the angle across from side a, and B is the angle across from side b. In this case, our proportion is set up like this:

\(\displaystyle \frac{7}{sin(40)}=\frac{8}{sin(x)}\) cross-multiply

\(\displaystyle 7sin(x)=8sin(40)\) evaluate the right side using a calculator

\(\displaystyle 7sin(x)= 5.14\) divide both sides by 7

\(\displaystyle sin(x)=0.734\) solve for x by evaluating \(\displaystyle sin^{-1}(0.734)\) in a calculator

\(\displaystyle x = 47.25^o\)

There is another solution as well. If \(\displaystyle 47.25^o\) has a sine of 0.734, so will its supplementary angle, \(\displaystyle 132.75^o\). 

Since \(\displaystyle 132.75^o + 40^o\) is still less than \(\displaystyle 180^o\), \(\displaystyle 132.75^o\) is a possible value for x.

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From \(\displaystyle \frac{\sin A}{a}=\frac{\sin B}{b}\), we get \(\displaystyle b\cdot \frac{\sin A}{a}=\sin B\). In this equation, if \(\displaystyle \sin B > 1\), no angle A that satisfies the triangle can be found. If \(\displaystyle \sin B = 1\), \(\displaystyle B=90^\circ\) and there is a right triangle determined. Finally, if \(\displaystyle \sin B < 1\), two measures of angle B can be calculated:  an acute angle B and an obtuse angle \(\displaystyle B'=180^\circ-B\). In this case, there may be one or two triangles determined. If \(\displaystyle B'+A\geq 180^\circ\), then the angle B' is not a solution.

In this problem, \(\displaystyle \sin B = 1\), \(\displaystyle B=90^\circ\) and there is one right triangle determined. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

\(\displaystyle \small \frac{b}{\sin\angle B} = \frac{a}{\sin \angle A}\)

Inputting the lengths of the triangle into this equation

\(\displaystyle \small \frac{8}{\sin(90^o)} = \frac{7}{\sin\angle A}\)

Isolating \(\displaystyle \small \angle A\)

\(\displaystyle \small \angle A = \sin^{-1}(\frac{7\sin(90^o)}{8})\)

\(\displaystyle \small \angle A = 61^o\)

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From \(\displaystyle \frac{\sin A}{a}=\frac{\sin B}{b}\), we get \(\displaystyle b\cdot \frac{\sin A}{a}=\sin B\). In this equation, if \(\displaystyle \sin B > 1\), no \(\displaystyle \angle A\) that satisfies the triangle can be found. If \(\displaystyle \sin B = 1\), \(\displaystyle B=90^\circ\) and there is a right triangle determined. Finally, if \(\displaystyle \sin B < 1\), two measures of \(\displaystyle \angle B\) can be calculated: an acute \(\displaystyle \angle B\) and an obtuse \(\displaystyle \angle\)\(\displaystyle B'=180^\circ-B\). In this case, there may be one or two triangles determined. If \(\displaystyle B'+A\geq 180^\circ\), then the \(\displaystyle \angle B'\) is not a solution.

In this problem, \(\displaystyle \sin\angle A=.087< 1\), so there may be one or two angles that satisfy this triangle. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

\(\displaystyle \small \frac{b}{\sin\angle B} = \frac{a}{\sin\angle A}\)

Inputting the values from the problem

\(\displaystyle \small \frac{10}{\sin\angle B} = \frac{4}{\sin\angle 5^o}\)

\(\displaystyle \small \angle B = \sin^{-1}(\frac{10\sin\angle 5^o}{4})\)

\(\displaystyle \small \small \angle B = 12.6^o\)

When the original given angle (\(\displaystyle \angle A=5^\circ\)) is acute, there will be:

• One solution if the side opposite the given angle is equal to or greater than the other given side
• No solution, one solution (right triangle), or two solutions if the side opposite the given angle is less than the other given side

In this problem, the side opposite the given angle is \(\displaystyle a=4\), which is less than the other given side \(\displaystyle b=10\). Therefore, we have a second solution. Find it by following the below steps:

\(\displaystyle \angle B'=180^\circ-B=180^\circ-12.6^\circ=167.4^\circ\)

\(\displaystyle \angle B'+\angle A=167.4^\circ+5^\circ=172.4^\circ\)\(\displaystyle \leq180^\circ\), so \(\displaystyle \angle B'\) is a solution.

Therefore there are two values for an angle, \(\displaystyle \angle B=12.6^\circ\)and \(\displaystyle \angle B'=167.4^\circ\).