This section covers:
Review of Right Triangle Trig
We learned about Right Triangle Trigonometry here, where we could “solve” triangles to find missing pieces (angles or sides).
Here is a review of the basic trigonometric functions, shown with both the SOHCAHTOA and Coordinate System Methods. Note that the second set of three trig functions are just the reciprocals of the first three; this makes it a little easier!
Remember that the sin (cos, and so on) of an angle is just a number!
Right Triangle 
SOHCAHTOA Method 
Coordinate System Method 
\(\displaystyle \begin{align}\text{SOH: Sine}\left( A \right)=\sin \left( A \right)=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}\\\text{CAH: Cosine}\left( A \right)=\cos \left( A \right)=\frac{{\text{Adjacent}}}{{\text{Hypotenuse}}}\\\text{TOA: Tangent}\left( A \right)=\tan \left( A \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}\end{align}\)
\(\displaystyle \begin{align}\text{cosecant}\left( A \right)=\csc \left( A \right)=\frac{1}{{\sin \left( A \right)}}=\frac{{\text{ Hypotenuse}}}{{\text{Opposite}}}\\\text{secant}\left( A \right)=\sec \left( A \right)=\frac{1}{{\cos \left( A \right)}}=\frac{{\text{ Hypotenuse}}}{{\text{Adjacent}}}\\\text{cotangent}\left( A \right)=\cot \left( A \right)=\frac{1}{{\tan \left( A \right)}}=\frac{{\text{ Adjacent}}}{{\text{Opposite}}}\end{align}\) 
\(\displaystyle \begin{align}\sin \left( A \right)=\frac{y}{h}\\\cos \left( A \right)=\frac{x}{h}\\\tan \left( A \right)=\frac{y}{x}\end{align}\)
\(\displaystyle \begin{align}\csc \left( A \right)=\frac{1}{{\sin \left( A \right)}}=\frac{h}{y}\\\sec \left( A \right)=\frac{1}{{\cos \left( A \right)}}=\frac{h}{x}\\\cot \left( A \right)=\frac{1}{{\tan \left( A \right)}}=\frac{x}{y}\end{align}\) 
We use the Law of Sines and Law of Cosines to “solve” triangles (find missing angles and sides) when we do not have a right triangle (which is called an oblique triangle). This is a little more complicated, and we have to know which angles and sides we do have to know which Law to use, but it’s not too bad.
Note that the Law of Sines can still be used to solve Right Triangles, using the 90° angle as one of the angles. It just turns out that the sin of 90° is 1; so it turns into a SOH case.
Law of Sines
The Law of Sines (or Sine Rule) provides a simple way to set up proportions to get other parts of a triangle that isn’t necessarily a right triangle.
We use the Law of Sines when we have the following parts of a triangle, as shown below: Angle, Angle, Side (AAS), Angle, Side, Angle (ASA), and Side, Side, Angle (SSA). Note that it won’t work when we only know the Side, Side, Side (SSS) or the Side, Angle, Side (SAS) pieces of a triangle. (Remember that these are “in a row” or adjacent parts of the triangle).
The only problem is that sometimes, with the SSA case, depending on what we know about the other sides and angles of the triangle, the triangle could actually have two different shapes (one acute and one obtuse). For these case, we have to account for both those shapes (so we’ll basically have two answers for the triangle, or maybe even no triangle can be formed). This is called the Ambiguous Case, and we’ll discuss it below here.
Once we know the formula for the Law of Sines, we can look at a triangle and see if we have enough information to “solve” it. “Solving a triangle” means finding any unknown sides and angles for that triangle (there should be six total for each individual triangle).
Note that we usually depict angles in capital letters, and the sides directly across from them in the same letter, but in lower case; see leftmost column for Law of Sines:
Law of Sines  Law of Cosines 
Use Law of Sines when you have these parts of a Triangle in a row:
^{*}This is where we have to look for the ambiguous case – remember “bad” word. 
Use Law of Cosines when you have these parts of a Triangle in a row:
One way I help remember the Law of Cosines is that the variable on the left side (for example,\({{a}^{2}}\) ) is the same as the angle variable (for example \(\cos A\)), and the other two variables (for example, \(b\) and \(c\)) are in the rest of the equation. 
Note: When using the Law of Cosines to solve the whole triangle (all angles and sides), particularly in the case of an obtuse triangle, you have to either finish solving the whole triangle using Law of Cosines (which is typically more difficult), or use the Law of Sines starting with the next smallest angle (the angle across from the smallest side) first. This is because of another case of ambiguous triangles.

Let’s do some problems; let’s first use the Law of Sines to find the indicated side or angle.
Remember that if we can’t solve it from what we have, and we have two of the three angles, we can obtain the third angle from Geometry (the sum of angles in a triangle is 180°).
Note that the triangles aren’t typically drawn to scale, meaning the angles and side measurements don’t exactly match the pictures. When I draw the triangles, I typically put the A and B angles on the “ground”.
When you try to draw the triangles to scale, you’ll see that larger angles are opposite larger sides, and smaller angles are opposite smaller sides. This could be a way you can check to see if you’re getting the correct answers.
And don’t forget to put your calculator in “DEGREE” mode.
I tend to round the angle measurements to a tenth of a degree, and the side measurements two decimal places (hundredths). If you want more accurate measurements when you start calculating the other sides and angles, you can use the STO> function in your calculator, or retype the long decimals you may get. To do this, use “STO>X” after seeing the value you want stored, and then “X” when you want to retrieve that value.
Using Law of Sines Formula to Find Indicated Parts of Triangle  
Solve for a: \(\displaystyle \frac{{\sin A}}{a}=\frac{{\sin B}}{b}=\frac{{\sin C}}{c}\)
We have an AAS case, so we have: \(\displaystyle \frac{{\sin \left( {44} \right)}}{{12}}=\frac{{\sin \left( {73} \right)}}{a}\)
Cross multiply (put calculator in DEGREE mode) to get: \(\displaystyle a=\frac{{\sin \left( {73} \right)\cdot 12}}{{\sin \left( {44} \right)}}=16.52\) 
Solve for A: \(\displaystyle \frac{{\sin A}}{a}=\frac{{\sin B}}{b}=\frac{{\sin C}}{c}\)
We have an SSA case that happens not to be the ambiguous case. (We’ll see below how to determine this). So, we have: \(\displaystyle \frac{{\sin A}}{{25}}=\frac{{\sin \left( {140} \right)}}{{45}}\) Cross multiply (put calculator in DEGREE mode and use 2^{nd} SIN for \({{\sin }^{{1}}}\)) to get: \(\displaystyle A={{\sin }^{{1}}}\left( {\frac{{\sin \left( {140} \right)\cdot 25}}{{45}}} \right)=20.9{}^\circ \) Here’s how we can put this in the graphing calculator with DEGREE mode: 
Solve for B: \(\displaystyle \frac{{\sin A}}{a}=\frac{{\sin B}}{b}=\frac{{\sin C}}{c}\)
We have an ASA case, but we need to find the missing angle C by knowing that the sum of angles in a triangle is 180°; the missing angle is \(180(63+74)=43\)°.
Now we have: \(\displaystyle \frac{{\sin \left( {43} \right)}}{{42.2}}=\frac{{\sin \left( {74} \right)}}{b}\)
Cross multiply (put calculator in DEGREE mode) to get: \(\displaystyle b=\frac{{\sin \left( {74} \right)\cdot 42.2}}{{\sin \left( {43} \right)}}=59.48\) 
Here are some problems where we need to “solve” the triangle, using the Law of Sines. Again, solving the triangle means finding all the missing parts, both sides and angles.
Using Law of Sines Formula to Solve Triangles  
\(m\angle A=25{}^\circ ,m\angle B=105{}^\circ ,a=450\)
Let’s first draw the triangle. Note that we can get the measurement of Angle C since \(25+105+50=180°\): \(\displaystyle \frac{{\sin A}}{a}=\frac{{\sin B}}{b}=\frac{{\sin C}}{c}\)
We have an AAS case, and we have: \(\displaystyle \frac{{\sin \left( {25} \right)}}{{450}}=\frac{{\sin \left( {105} \right)}}{b}=\frac{{\sin \left( {50} \right)}}{c}\)
Solving for b and c, we get:

\(m\angle A=30{}^\circ ,m\angle B=110{}^\circ ,c=12.5\)
Let’s first draw the triangle. Note that we can get the measurement of Angle C since \(30+110+40=180°\): \(\displaystyle \frac{{\sin A}}{a}=\frac{{\sin B}}{b}=\frac{{\sin C}}{c}\)
We have an ASA case, and we have: \(\displaystyle \frac{{\sin \left( {30} \right)}}{a}=\frac{{\sin \left( {110} \right)}}{b}=\frac{{\sin \left( {40} \right)}}{{12.5}}\)
Solving for a and b, we get:

\(a=22,\,\,\,b=14,\,\,\,\,m\angle A=110{}^\circ \)
Let’s first draw the triangle: \(\displaystyle \frac{{\sin A}}{a}=\frac{{\sin B}}{b}=\frac{{\sin C}}{c}\)
We have an SSA (nonambiguous) case, and we have: \(\displaystyle \frac{{\sin \left( {110} \right)}}{{22}}=\frac{{\sin \left( B \right)}}{{14}}=\frac{{\sin \left( C \right)}}{c}\)
Solve first for angle B (crossmultiply and use 2^{nd} SIN), to get B = 36.7°. Then we can get C: \(180(110+36.7)=33.3°\). Now we can complete the table:

Law of Sines Ambiguous Case (SSA)
When we have the SideSideAngle (in a row) case (SSA), we could have one, two, or no triangles formed, and we have to do extra work to determine which situation we have.
In these cases, I always like to draw my triangle with the known angle on the bottom left (even if that angle is B or C), so I can see what’s going on. If the side directly across from this angle (the paired side) is less than the side touching this angle, we will probably have an ambiguous case (or may have no triangle that can be formed).
In this situation, if we get an error message in the calculator when trying to get the other angle using Law of Sines, or, in the case of an obtuse triangle, we get more than 180° for the triangle, there is no triangle that can be formed with the numbers given (and that’s the answer).
For acute triangles, if we get an answer other than 90°, we will have two triangles that can be formed, and the second angle is 180 minus the angle we just got (one will be acute and one will be obtuse). We can then solve for two different triangles (the given two sides and one angle for the two triangles will be the same). If we do get 90° for the second angle, we have one right triangle. This happens when the height of the triangle equals the paired side (the side across from the known angle).
Here’s an illustration of this:
Law of Sines Ambiguous Case: Given: a, b, and A (SSA) 
For obtuse triangles, we’ll have either one triangle (when \(a>b\)) or no triangle (when \(a\le b\)). For acute triangles: If \(a\ge b\), we are OK: one triangle! Proceed as we did above with Law of Sines: \(\displaystyle \frac{{\sin A}}{a}=\frac{{\sin B}}{b}=\frac{{\sin C}}{c}\).
If \(a<b\), we have the ambiguous case; we either have two triangles (if \(a>\) the height \(h\) of the triangle), one triangle (if \(a=h\)), or no triangle (if \(a<h\)). (Note that if there is no triangle, we’ll get an error on our calculator when trying to solve for the angle \(B\).)
(Note that we can get the height by using right triangle trig: the height is \(h=b\sin A\) since \(\displaystyle \sin A=\frac{h}{a}\). You don’t usually have to get the height though, unless you are asked to; your calculator will get an angle other than 90° without an error if there are two triangles.) If \(a=h\), we have a right triangle, where angle \(B=90°\).
If \(a>h\), we can form two triangles with side \(b\): one with sides \(a\) (to the right above) and \({{c}_{1}}\), and the other with sides \(a\) (to the left above) and \({{c}_{2}}\). Then angles \({{B}_{1}}\) and \({{C}_{1}}\) will be associated with \(a\) and \({{c}_{1}}\), and angles \({{B}_{2}}\) and \({{C}_{2}}\) will be associated with \(a\) and \({{c}_{2}}\), and we use the formulas below. To get \({{B}_{2}}\), we simply subtract \({{B}_{1}}\) from 180°: \({{B}_{2}}=180{{B}_{1}}\).
\(\displaystyle \frac{{\sin A}}{a}=\frac{{\sin {{B}_{1}}}}{b}=\frac{{\sin {{C}_{1}}}}{{{{c}_{1}}}}\) and \(\displaystyle \frac{{\sin A}}{a}=\frac{{\sin {{B}_{2}}}}{b}=\frac{{\sin {{C}_{2}}}}{{{{c}_{2}}}}\) . Now we’ll have all 6 parts for each of the two triangles, with \(a, b\), and \(A\) being the same for both. 
Let’s do some problems, so it won’t seem so confusing. Solve for all possible triangles with the given conditions:
Law of Sines SSA Ambiguous Case Problems – Solving Triangles  
\(b=25,\,\,c=20\,,\,\,m\angle C=130{}^\circ \)
Let’s first draw the triangle (which is obviously not to scale):
Since this is an obtuse triangle, and since \(20<25\), we won’t have a triangle. But let’s see what happens when using Law of Sines:
\(\displaystyle \frac{{\sin \left( A \right)}}{a}=\frac{{\sin \left( B \right)}}{{25}}=\frac{{\sin \left( {130} \right)}}{{20}}\,\,\)
Solve first for angle B (crossmultiply and use 2^{nd} SIN), to get \(B=73.2{}^\circ \). Then we can get A: \(180\left( {130+73.2} \right)=23.2{}^\circ \). Aha! We can’t have a negative angle! No triangle exists!
Note: If the side (\(c\)) across from Angle C was 20, for example (less than \(b\), which is 25), we’d have one triangle and could solve the normal way using Law of Sines. 
\(b=14,\,\,\,c=20,\,\,\,\,m\angle B=40{}^\circ \)
Let’s first draw the triangle: Since \(14<20\), we have the ambiguous case; we either have two triangles (if \(14>\) the height \(h\) of the triangle), one triangle (if \(14=h\)), or no triangle (if \(14<h\)). Let’s solve for angle C to see whether or not we get an error message when we take \({{\sin }^{{1}}}\): \(\displaystyle \frac{{\sin \left( A \right)}}{a}=\frac{{\sin \left( {40} \right)}}{{14}}=\frac{{\sin \left( C \right)}}{{20}}\)
Crossmultiply to get C (using 2^{nd} SIN) to get \(C=66.7{}^\circ \). Since we didn’t get an error, or an angle measurement of 90°, we have two triangles, one with \(C=66.7°\), and the other with \(C= 180°–66.7°=113.3°\).
Now we can get Angle A, and solve for side a in both cases. Note how the original values (B, b, and c) stay the same in both cases: Triangle 1:
Triangle 2:

\(a=30,\,\,\,b=80,\,\,\,\,m\angle A=50{}^\circ \)
Let’s first draw the triangle; it will actually end up looking something like this:
Since \(a<b\), we have the ambiguous case; we either have two triangles (if \(a>\) the height \(h\) of the triangle), one triangle (if \(a=h\)), or no triangle (if \(a<h\)).
Let’s try to solve for angle B and see that we do get an error message when we use \({{\sin }^{{1}}}\) (we get a \(\sin \left( B \right)\) of greater than 1):
\(\displaystyle \frac{{\sin \left( {50} \right)}}{{30}}=\frac{{\sin \left( B \right)}}{{80}}=\frac{{\sin \left( C \right)}}{c}\)
And as it turns out, the height of this triangle is \(h=b\sin A=80\sin \left( {50} \right)=61.3\), which is \(>30\). Since \(a<h\), no triangle can be formed with what we are given. 
Note: We can also solve ambiguous case triangles using the Law of Cosines and a graphing calculator here).
Here’s another type of problem you might encounter when learning about the Law of Sines Ambiguous Case:
Ambiguous Case Problem  Solution 
Given a triangle ABC, what values of a would result in making two triangles if \(A = 40°\) and \(b=10\)?  In order for there to be two triangles, a must be less than 10, but greater than the height of the triangle, which is \(10\sin 40{}^\circ \), or 6.427.
Therefore, \(6.427<a<10\). 
Law of Cosines
The Law of Cosines (or Cosine Rule) again provides a simple way to set up proportions to get other parts of a triangle that isn’t necessarily a right triangle.
We use the Law of Cosines when we have the following parts of a triangle, as shown below: Side, Angle, Side (SAS), and Side, Side Side (SSS). It also will work for the Side, Side, Angle (SSA) case, and we will see that here, but the Law of Sines is usually taught with this case, because of the Ambiguous Case.
(Remember that these are “in a row” or adjacent parts of the triangle).
Once we know the formula for the Law of Cosines, we can look at a triangle and see if we have enough information to “solve” it. “Solving a triangle” means finding any unknown sides and angles for that triangle (there should be six total for each individual triangle).
Again, note that we usually depict angles in capital letters, and the sides directly across from them in the same letter, but in lower case; see rightmost column for Law of Cosines:
Law of Sines  Law of Cosines 
Use Law of Sines when you have these parts of a Triangle in a row:
^{*}This is where we have to look for the ambiguous case – remember “bad” word. 
Use Law of Cosines when you have these parts of a Triangle in a row:
One way I help remember the Law of Cosines is that the variable on the left side (for example,\({{a}^{2}}\) ) is the same as the angle variable (for example \(\cos A\)), and the other two variables (for example, \(b\) and \(c\)) are in the rest of the equation. 
Note: When using the Law of Cosines to solve the whole triangle (all angles and sides), particularly in the case of an obtuse triangle, you have to either finish solving the whole triangle using Law of Cosines (which is typically more difficult), or use the Law of Sines starting with the next smallest angle (the angle across from the smallest side) first. This is because of another case of ambiguous triangles.

Let’s do some problems; let’s first use the Law of Cosines to find the indicated side or angle.
Remember again that the triangles aren’t typically drawn to scale, meaning the angles and side measurements don’t exactly match the pictures. When I draw the triangles, I typically put the A and B angles on the “ground”.
Again, notice that if you try to draw the triangles to scale, you’ll see that larger angles are opposite larger sides, and smaller angles are opposite smaller sides. This could be a way you can check to see if you’re getting the correct answers.
And don’t forget to put your calculator in “DEGREE” mode.
Using Law of Cosines Formula to Find Indicated Parts of Triangle  
Solve for a:
\(\displaystyle {{a}^{{2\,}}}={{b}^{2}}+{{c}^{2}}2bc\cos A\)
We have an SAS case; let’s solve for \(a\):
\(\displaystyle \begin{array}{c}{{a}^{{2\,}}}={{16}^{2}}+{{14}^{2}}2\left( {16} \right)\left( {14} \right)\cos \left( {100} \right)\\a=\sqrt{{256+196\left( {448} \right)\left( {.173648} \right)}}\\=23.02\end{array}\)
Note that angle A is called the included angle between sides b and c, since it is located between the two sides.

Solve for A:
\(\displaystyle {{a}^{{2\,}}}={{b}^{2}}+{{c}^{2}}2bc\cos A\)
We have an SSS case; let’s solve for A:
\(\displaystyle \begin{array}{c}{{14.5}^{2}}={{10.5}^{2}}+{{12.2}^{2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {10.5} \right)\left( {12.2} \right)\cos \left( A \right)\\{{14.5}^{2}}{{10.5}^{2}}+{{12.2}^{2}}=\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {10.5} \right)\left( {12.2} \right)\cos \left( A \right)\\\cos \left( A \right)=\frac{{{{{14.5}}^{2}}{{{10.5}}^{2}}{{{12.2}}^{2}}}}{{2\left( {10.5} \right)\left( {12.2} \right)}}\\A={{\cos }^{{1}}}\left( {\frac{{{{{14.5}}^{2}}{{{10.5}}^{2}}{{{12.2}}^{2}}}}{{2\left( {10.5} \right)\left( {12.2} \right)}}} \right)=79.0{}^\circ \end{array}\) 
Solve for B:
\(\displaystyle {{b}^{{2\,}}}={{a}^{2}}+{{c}^{2}}2ac\cos B\)
We have an SSS case; let’s solve for \(B\):
\(\displaystyle \begin{align}{{6}^{2}}&={{12}^{2}}+{{5}^{2}}2\left( {12} \right)\left( 5 \right)\cos \left( B \right)\\B&={{\cos }^{{1}}}\left( {\frac{{{{6}^{2}}{{{12}}^{2}}{{5}^{2}}}}{{2\left( {12} \right)\left( 5 \right)}}} \right)\\&=ERROR\end{align}\)
No triangle can be formed with these side lengths. Note that this case doesn’t work because the sum of any two sides of a triangle must be greater than the third side: \(\require{cancel} 6+5\cancel{\ge }\,\,12\). 
Note: Here’s how we can put the first problem above in the graphing calculator (DEGREE mode) to solve:
Here’s how we can put the second problem above in the calculator (DEGREE mode). Be sure to use the “– ” (minus) under the “\(x\)” for the first part, and negative sign (“–”) for the second:
Or, in three steps, using “2^{nd} COS 2^{nd} ()” in the last step:
Or, using the fraction feature on the calculator, we can put in “2^{nd} COS ALPHA Y= ENTER” to get a fraction format, and then enter the rest:
Now let’s solve for all possible triangles with the given conditions.
Caution: When using the Law of Cosines to solve the whole triangle (all angles and sides), particularly in the case of an obtuse triangle, you have to either finish solving the whole triangle using Law of Cosines (which is typically more difficult), or use the Law of Sines starting with the next smallest angle (the angle across from the smallest side) first. This is because of another example of ambiguous cases with triangles.
Using Law of Cosines Formula to Solve Triangles  
\(a=3,\,\,\,b=5,\,\,\,m\angle C=40{}^\circ \) \(\displaystyle {{c}^{{2\,}}}={{a}^{2}}+{{b}^{2}}2ab\cos C\)
We have an SAS case; let’s solve for \(c\):
\(\displaystyle \begin{align}{{c}^{{2\,}}}&={{3}^{2}}+{{5}^{2}}2\left( 3 \right)\left( 5 \right)\cos \left( {40} \right)\\c&=\sqrt{{9+25\left( {30} \right)\left( {.766044} \right)}}\\&=3.319\end{align}\)
Now we can either use Law of Cosines again to gets angles A or B, or use Law of Sines for angle A first, since \(\displaystyle \text{angle }A<\text{angle }B\) (since \(\displaystyle \text{side }a<\text{side }b\)); this is easier:
\(\displaystyle \frac{{\sin A}}{3}=\frac{{\sin \left( {40} \right)}}{{3.319}};\,\,\,A=35.5{}^\circ \)
Now we can subtract the sum of angles A and C from 180° to get angle B:

\(a=30,\,\,\,b=35,\,\,\,c=25\) \(\displaystyle {{c}^{{2\,}}}={{a}^{2}}+{{b}^{2}}2ab\cos C\)
We have an SSS case; let’s solve for A first:
\(\displaystyle \begin{align}{{30}^{{2\,}}}&={{35}^{2}}+{{25}^{2}}2\left( {35} \right)\left( {25} \right)\cos \left( A \right)\\A&={{\cos }^{{1}}}\left( {\frac{{{{{30}}^{2}}{{{35}}^{2}}{{{25}}^{2}}}}{{2\left( {35} \right)\left( {25} \right)}}} \right)\\&=57.1{}^\circ \end{align}\)
Now we can either use Law of Cosines again to gets angles B or C, or use Law of Sines for angle C first, since \(\displaystyle \text{angle }C<\text{angle }B\) (since \(\displaystyle \text{side }c<\text{side }b\)); this is easier:
\(\displaystyle \frac{{\sin \left( {57.1} \right)}}{{30}}=\frac{{\sin C}}{{25}};\,\,\,C=44.4{}^\circ \)
Now we can subtract the sum of angles A and C from 180° to get angle B:

Here’s a tricky one where we need to use the Law of Cosines twice:
Law of Cosines Problem  Solution and Explanation 
Find \(\overline{{BD}}\):  First, use Law of Cosines on the large triangle to get \(\angle C\):
\(\displaystyle \begin{align}{{c}^{{2\,}}}&={{a}^{2}}+{{b}^{2}}2ab\cos C\\{{12}^{2}}&={{8}^{2}}+{{10}^{2}}2\left( 8 \right)\left( {10} \right)\cos C\\C&={{\cos }^{{1}}}\left( {\frac{{{{{12}}^{2}}{{8}^{2}}{{{10}}^{2}}}}{{2\left( 8 \right)\left( {10} \right)}}} \right)\\C&={{\cos }^{{1}}}\left( {\frac{1}{8}} \right)=82.82{}^\circ \end{align}\) Then, use the Law of Cosines on the smaller triangle (containing \(\angle C\)) to get \(\overline{{BD}}\): \(\displaystyle \begin{align}{{\left( {\overline{{BD}}} \right)}^{2}}&={{8}^{2}}+{{4}^{2}}2\left( 8 \right)\left( 4 \right)\cos C\\{{\left( {\overline{{BD}}} \right)}^{2}}&=64+1664\cos \left( {{{{\cos }}^{{1}}}\left( {\frac{1}{8}} \right)} \right)\\{{\left( {\overline{{BD}}} \right)}^{2}}&=64+1664\left( {\frac{1}{8}} \right)=72\\\overline{{BD}}&=\sqrt{{72}}=6\sqrt{2}\end{align}\) 
Law of Cosines Ambiguous Case (SSA)
Let’s solve an SSA Ambiguous Case Triangle using the Law of Cosines instead of the Law of Sines. We can do this fairly easily using a graphing calculator; in fact the calculator can actually tell us how many triangles we will get! We’ll use the same problem that we used earlier.
Law of Cosines SSA Ambiguous Case Problem using Graphing Calculator  
\(a=14,\,\,\,b=20,\,\,\,\,m\angle A=40{}^\circ \) Let’s first draw the triangle: Since \(a<b\), we have the ambiguous case; we either have two triangles (if \(a>\) the height \(h\) of the triangle), one triangle (if \(a=h\)), or no triangle (if \(a<h\)). Even though we typically use the Law of Sines for the ambiguous case, let’s use the Law of Cosines: \(\displaystyle {{a}^{{2\,}}}={{b}^{2}}+{{c}^{2}}2bc\cos A:\,\,\,\,\,\,\,\,\,{{14}^{{2\,}}}={{20}^{2}}+{{c}^{2}}2\left( {20} \right)\left( c \right)\cos \left( {40{}^\circ } \right)\). Since it’s a quadratic, we’ll either get one answer for c (if one triangle can be formed), no answer (if no triangle can be formed), or two answers (if two triangles can be formed). In this case, we’ll see that two triangles can be formed since we’ll get two different answers (intersections) for c. The trick is to put this equation in a graphing calculator and solve for c using \({{Y}_{1}}\) and \({{Y}_{2}}\) and the Intersect function (you might need to use ZOOM 6, ZOOM 0, and then ZOOM 3, ENTER a few times to see the intersection, and then use TRACE and move cursor closer to the second intersection): Now that we have the two answers for c, we can use the Law of Sines to solve for the two solutions for angle C: \(\displaystyle \frac{{\sin \left( {40} \right)}}{{14}}=\frac{{\sin C}}{{9.78}};\,\,\,C=26.7{}^\circ \) \(\displaystyle \frac{{\sin 40}}{{14}}\,\,=\,\,\frac{{\sin C}}{{20.86}};\,\,\,C=73.3{}^\circ \) Now we can get Angle B for each of the cases (by subtracting angles A and C from 180°), and we have solved both triangles: 

Triangle 1:

Triangle 2:

Areas of Triangles
In Geometry we learned that we can get the area of triangles quite easily if we know the base of the triangle and the altitude (which is a line that is perpendicular to the base and extends up to the top of the triangle): \(\displaystyle \text{Area}=\frac{1}{2}bh\) or \(\displaystyle \text{Area}=\frac{{bh}}{2}\).
Now that we know trig, we get the area of a triangle without having to know the altitude if we know two sides, and the angle inside the two sides (the SideAngleSide or SAS case), or three sides of the triangle (SideSideSide, or SSS case).
And remember that many times if you don’t initially have these cases, you can solve the triangle using the Law of Sines or Law of Cosines to get these angles and sides.
Here are the two formulas; note that one uses the sin function, and the other (Heron’s formula) uses the sides only, with one half the sum of the sides being what is called a semiperimeter:
Triangle  Area Formula  Use this Area Formula when you have these parts of a Triangle in a row: 
Area of Triangle:
\(\displaystyle \begin{array}{l}\text{Area}=\frac{1}{2}\left( b \right)\left( c \right)\left( {\sin A} \right)\\\text{Area}=\frac{1}{2}\left( a \right)\left( c \right)\left( {\sin B} \right)\\\text{Area}=\frac{1}{2}\left( a \right)\left( b \right)\left( {\sin C} \right)\end{array}\)


Area of Triangle (Heron’s):
\(\displaystyle \text{Area}=\sqrt{{s\left( {sa} \right)\left( {sb} \right)\left( {sc} \right)}}\), where \(\displaystyle s=\frac{1}{2}\left( {a+b+c} \right)\) 

Finding Area of Triangles  
\(a=3,\,\,\,b=5,\,\,\,m\angle C=40{}^\circ \) \(\displaystyle \begin{align}\text{Area}=\frac{1}{2}\left( a \right)\left( b \right)\left( {\sin C} \right)\end{align}\)
We have an SAS case; let’s get the area of this triangle: \(\displaystyle \begin{align}\text{Area}&=\frac{1}{2}\left( a \right)\left( b \right)\left( {\sin C} \right)\\&=\frac{1}{2}\left( 3 \right)\left( 5 \right)\sin \left( {40} \right)=4.82\end{align}\)
Here’s another where we have to use Law of Sines first, since we don’t have the SAS case: \(b=100,\,\,\,c=75,\,\,\,m\angle C=30{}^\circ \) \(\displaystyle \frac{{\sin \left( {30} \right)}}{{75}}=\frac{{\sin \left( B \right)}}{{100}};\,\,\,\,\,B=41.8{}^\circ ; \,\,\,A=108.2{}^\circ \) \(\displaystyle \begin{align}\text{Area}&=\frac{1}{2}\left( b \right)\left( c \right)\left( {\sin A} \right)\\&=\frac{1}{2}\left( {100} \right)\left( {75} \right)\sin \left( {108.2} \right)=3562.395\end{align}\) 
\(a=30,\,\,\,b=35,\,\,\,c=25\) \(\displaystyle \text{Area}=\sqrt{{s\left( {sa} \right)\left( {sb} \right)\left( {sc} \right)}}\), where \(\displaystyle s=\frac{1}{2}\left( {a+b+c} \right)\)
We have an SSS case; let’s get the area of the triangle by first finding \(\displaystyle s:\,\,\,\,\,\,s=\frac{1}{2}\left( {30+35+25} \right)=45\). Then: \(\displaystyle \begin{align}\text{Area}&=\sqrt{{s\left( {sa} \right)\left( {sb} \right)\left( {sc} \right)}}\\&=\sqrt{{45\left( {4530} \right)\left( {4535} \right)\left( {4525} \right)}}\\&=\sqrt{{45\left( {15} \right)\left( {10} \right)\left( {20} \right)}}\\&=367.42\end{align}\)
A little bit more complicated, but not too bad! 
Applications/Word Problems
Here are some examples of applications of the Law of Sines, Law of Cosines, and Area of Triangles.
One big hint on doing these word problems is to try to draw the diagram (if they don’t give you one) as much to scale as possible, so you can see if your answers make sense! For example, draw the angles as close to the correct angle measurements and sides in the proportion of the numbers they give you.
Here’s an example of how we might use the Law of Sines to get distances that are typically difficult to measure. This stuff is really used in “real life”!
Law of Sines Problem  Solution and Explanation 
Ali and Brynn are standing 250 yards apart. Both girls sight an airplane with angles of elevation 40° and 45°, respectively. How far from the plane is Ali?
First, draw a picture: 
To get the distance \(b\) from Ali (left) to the plane, we first get the measurement of the third angle, which is 95°, and then use the Law of Sines:
\(\displaystyle \frac{{\sin \left( {95} \right)}}{{250}}=\frac{{\sin \left( {45} \right)}}{b}\)
Cross multiply to get \(b=177.45\).
The distance from Ali to the plane is roughly 177.45 yards. 
Here are a few problems that use the Law of Cosines for a parallelogram (remember, from Geometry?) By definition, a parallelogram is a quadrilateral (foursided figure with straight sides) that has opposite parallel sides, and it turns out that opposite sides are equal. Parallel means never crossing, like railroad tracks. For a parallelogram ABCD, we need to draw ABCD in a row (in any direction) around the figure.
Draw the parallelograms to see how to solve the problem. Remember that adjacent angles in a parallelogram add up to 180°. These are called Same Side Interior angles.
Law of Cosines Problem  Solution and Explanation 
As shown in the picture below, in a parallelogram \(ABCD\), \(AD=450\) feet, \(AB=240\) feet, and diagonal \(AC=290\) feet.
What is the measure of \(\angle BCD\)? Property of parallelogram: \(\angle ABC\text{ (}\angle \text{ }B\text{)}+\angle \text{ }BCD=180{}^\circ \) 
Draw the parallelogram starting with A in the upper right and going counterclockwise with ABCD (other ways will work, as long as you draw ABCD in a row). You can sort of see (and it’s true!) that \(AD=BC=450\); this is one of the properties of parallelograms: opposite sides are equal (congruent).
Using the Law of Cosines (SAS), we can get Angle B first: \(\displaystyle \begin{align}{{290}^{2}}&={{240}^{2}}+{{450}^{2}}2\left( {240} \right)\left( {450} \right)\cos \left( B \right)\\{{290}^{2}}{{240}^{2}}{{450}^{2}}&=2\left( {240} \right)\left( {450} \right)\cos \left( B \right)\\B&={{\cos }^{{1}}}\left( {\frac{{{{{290}}^{2}}{{{240}}^{2}}{{{450}}^{2}}}}{{2\left( {240} \right)\left( {450} \right)}}} \right)\\&\approx 35.4{}^\circ \end{align}\)
Now it just so happens that two adjacent angles (two angles in a row) of a parallelogram add up to 180°, so \(\angle BCD=180{}^\circ 35.4{}^\circ \approx 144.6{}^\circ \). 
Parallelogram has diagonals \(16\) and \(10\), with side \(\overline{{AD}}=7\).
What is \(\overline{{CD}}\)?

For another perspective, draw the parallelogram starting with A in the upper right and going clockwise with ABCD (other ways will work, as long as you draw ABCD in a row). Because of the figure is a parallelogram, we know that the diagonals bisect each other.
Use the Law of Cosines twice, first to get \(\angle AXD\) (indicated by \(X\)): \(\displaystyle \begin{align}{{7}^{2}}&={{8}^{2}}+{{5}^{2}}2\left( 8 \right)\left( 5 \right)\cos \left( X \right)\\{{7}^{2}}{{8}^{2}}{{5}^{2}}&=2\left( 8 \right)\left( 5 \right)\cos \left( X \right)\\X=&{{\cos }^{{1}}}\left( {\frac{{{{7}^{2}}{{8}^{2}}{{5}^{2}}}}{{2\left( 8 \right)\left( 5 \right)}}} \right)\\&=60{}^\circ \end{align}\) Use the Law of Cosines again to get \(\overline{{CD}}\), using the fact that \(\angle DXC=180{}^\circ 60{}^\circ =120{}^\circ \) (two adjacent angles, or a linear pair, are supplementary, or add up to \(180{}^\circ \)): \(\displaystyle \begin{align}{{\overline{{CD}}}^{2}}&={{8}^{2}}+{{5}^{2}}2\left( 8 \right)\left( 5 \right)\cos \left( {120} \right)\\{{\overline{{CD}}}^{2}}&=129\\\overline{{CD}}&\approx 11.358\end{align}\) 
This one’s a little tricky, since we have more than one triangle. (It’s actually a trapezoid, with two parallel sides and two nonparallel sides.) We’ll first work with the triangle where we have enough information, and then use a common side to solve the parts of the triangle we want.
Law of Sines/Cosines Problem  Solution and Explanation 
Three dogs are sitting in a kitchen and waiting to get their dog food. Dog A is 4.5 feet from Dog B, and Dog C is 2.5 feet from Dog A, as shown in the diagram below. Note that the three dogs’ positions form a right (90°) angle.
The angles formed at Dogs A and B to the dog food are 40° and 80°, respectively, as shown in the diagram.
How far is Dog C from the dog food?
First, draw a picture:

We can start out using the triangle between Dogs A and B and the dog food, where we have an ASA case (Law of Sines):
Let’s use Law of Sines to get the distance between Dog A and the food (side \(b\)). We have an ASA case, but we need to find the missing angle (60°) so we can get \(b\):
\(\displaystyle \frac{{\sin \left( {60} \right)}}{{4.5}}=\frac{{\sin \left( {80} \right)}}{b};\,\,\,\,\,\,\,\,b=5.1172\) (I used more decimal places for better accuracy).
Now we have an SAS case for the leftmost triangle, since we know that angle \(y=50°\) (the sum of 2 angles in a corner \(=90°\)). We can use the Law of Cosines to get side \(y\):
\(\displaystyle \begin{align}{{x}^{2}}&={{5.1172}^{2}}+{{2.5}^{2}}2\left( {5.1172} \right)\left( {2.5} \right)\cos \left( {50} \right)\\x&=\sqrt{{{{{5.1172}}^{2}}+{{{2.5}}^{2}}2\left( {5.1172} \right)\left( {2.5} \right)\cos \left( {50} \right)}}\\&=3.9987\approx 4\end{align}\)
The distance between Dog C and the dog food is about 4 feet.

Bearing Problem
(Note that this problem is actually solved more easily using vectors here in the Introduction to Vectors section.)
Let’s first talk about what it means to have a bearing of a certain degree, since this is typically used in navigation. First of all, think of north as going up (positive \(y\)axis), south as going down (negative \(y\)axis), east as going to the right (positive \(x\)axis), and west as going to the left (negative \(x\)axis).
Unless otherwise noted, bearing is the measure of the clockwise angle that starts due north or on the positive \(y\)axis (initial side) and terminates a certain number of degrees (terminal side) from that due north starting place. (This is also written, as in the case of a bearing of 40° as “40° east of north”, or “N40°E”).
Note: Sometimes, you’ll see a bearing that includes more directions, such as 70° west of north, also written as N70°W. In this case, the angle will start due north (straight up, or on the positive \(y\)axis) and go counterclockwise 70° (because it’s going west, or to the left, instead of east). Similarly, a bearing of 50° south of east, or E50°S, would be an angle that starts due east (on the positive \(x\)axis) and go clockwise 50° clockwise (towards the south, or down). Also, if you see a bearing of southwest, for example, the angle would be 45° south of west, or 225° clockwise from north, and so on.
And each time a boat or ship changes course, you have to draw another line to the north to map its new bearing.
Here are some bearing examples:
In this problem, we also have to remember that \(\text{Distance}=\text{Rate}\,\times \,\text{Time}\), since we are given rates and times and need to calculate distances.
Also, remember from Geometry that Alternate Interior Angles are congruent when a transversal cuts parallel lines.
Law of Cosines Problem  Solution and Explanation 
A cruise ship travels at a bearing of 40° (east of north) at 20 mph for 3 hours, and changes course to a bearing of 120° (east of north). It then travels 25 mph for 2 hours.
a) Find the distance the ship is from its original position. b) Find the ship’s new bearing from the original position. c) On what bearing must the ship travel to return back to its original position?
First, draw a picture:
Note that we get 60 miles by multiplying rate x time to get distance, or 20 mph · 3 hours. Similarly, we get 50 miles by multiplying 25 mph by 2 hours. 
Use Geometry to get angle B, which is \(40°+(180°–120)°=40°+60°=100°\) (the 40° is from Alternate Interior Angles).
a) Use the Law of Cosines to get \(b\), since we have an SAS case: \(\displaystyle \begin{align}{{b}^{2}}&={{a}^{2}}+{{c}^{2}}2\left( a \right)\left( c \right)\cos \left( B \right)\\{{b}^{2}}&={{50}^{2}}+{{60}^{2}}2\left( {50} \right)\left( {60} \right)\cos \left( {100} \right)\\b&=\sqrt{{{{{50}}^{2}}+{{{60}}^{2}}2\left( {50} \right)\left( {60} \right)\cos \left( {100} \right)}}\\b&\approx 84.51\end{align}\) The distance the ship is from its original position is 84.51 miles b) To find angle A, we can use Law of Cosines again (or Law of Sines, since it’s the smallest angle): \(\displaystyle \begin{align}{{a}^{2}}&={{b}^{2}}+{{c}^{2}}2\left( b \right)\left( c \right)\cos \left( A \right)\\{{50}^{2}}&={{84.51}^{2}}+{{60}^{2}}2\left( {84.51} \right)\left( {60} \right)\cos \left( A \right)\\\,A&={{\cos }^{{1}}}\left( {\frac{{{{{50}}^{2}}{{{84.51}}^{2}}{{{60}}^{2}}}}{{2\left( {84.51} \right)\left( {60} \right)}}} \right)\\A&\approx 35.6{}^\circ \end{align}\) \(\displaystyle \begin{align}\frac{{\sin \left( A \right)}}{a}&=\frac{{\sin \left( B \right)}}{b}\\\frac{{\sin \left( A \right)}}{{50}}&=\frac{{\sin \left( {100} \right)}}{{84.51}}\\A&\approx 35.6{}^\circ \end{align}\) The ship’s bearing from its original position is \(40°+35.6°= 75.6°\) (east of north). NOTE: In these types of problems, it’s also typical to have the ship end up in a different quadrant, such as the 4th quadrant (for example, if \(A\) were greater than 50° degrees). We would still just add the 40° to the angle we get to get the bearing. Try to draw the pictures as much to scale as you possibly can to see if your answer makes sense.
c) To get the bearing that the ship must travel to return back to its original position, add \(180°\) to the \(A+40{}^\circ \) to get \(180{}^\circ +35.6{}^\circ +40{}^\circ =255.6{}^\circ \) (east of north), which is the same as \(14.4°\) south of west.
This is what we got when we did this problem using Vectors here! 
Here’s one more bearing problem; note that there are probably many other ways of doing this problem, but this one works 😉:
Bearing Problem  Solution and Explanation 
Joa is standing 100 feet from her friend Rachel. From Joa, the bearing to Rachel is E25°S. From Joa to another friend, Emily, the bearing is S30°E. The bearing from Rachel to Emily is S80°W. What is the distance from Emily to Rachel?
Probably the most difficult part is to drawing a picture of the problem: 
We can first work with the bearing from Joa to Rachel, which is E25°S, or 25° south of east. Thus, we go straight east (to the right) of Joa, and then down 25° from there. Then we can draw a line from Joa to Emily, which is S30°E, or 30° east of south (we go straight south or down and then 30° east or to the right from there).
Now, because all the angles in a right angle (perfect corner) add up to 90° from Geometry, we can see that the angle inside the triangle near Joa is \(90°–(25+30)°=35°\).
Now let’s look at the angles close to Rachel. Since the bearing from Rachel to Emily is S80°W, we go straight down (south) from Rachel and 80° to west (to the left) to draw the line to Emily. For the angle inside the triangle, we know the top part of it is 25° by using a Geometry rule that Alternate Interior Angles in a transversal of parallel lines (horizontal lines) are congruent, and the second part of it by knowing that angles in a right angle (perfect corner) add up to 90°. Thus, the angles inside the triangle close to Rachel add up to 35°.
We can then get the angle close to Emily by subtracting 35° (Joa’s angle) and 35° (Rachel’s angle) from 180° to get 110°. Whew – now we’ve done the difficult part! Now we can use the Law of Sines to get \(x\), since we have an AAS (110°/35°/100 ft) case: \(\displaystyle \begin{align}\frac{{\sin \left( {110} \right)}}{{100}}&=\frac{{\sin \left( {35} \right)}}{x}\\x&=61.039\end{align}\) The distance from Emily to Rachel is about 61 feet. 
Here’s an Area Word Problem:
Area Problem  Solution and Explanation 
Jill, a surveyor, needs to approximate the area of a piece of land, as shown below. She walks the perimeter of the land and measures the side distances and one angle, as shown below. What is the area of the piece of land?
Note: Separate the piece of land into two triangles: 
Since we are working with triangles in this unit, we can see that we can make two triangles with the figure, and then start filling in the “pieces of the puzzle” with what we can solve. Let’s call the diagonal “\(x\)”; we need to solve for \(x\), so we can get the area of the lower triangle.
We need to start with the top triangle, using the Law of Cosines, since we have SAS:
\(\displaystyle \begin{align}{{x}^{2}}&={{45}^{2}}+{{25}^{2}}2\left( {45} \right)\left( {25} \right)\cos \left( {110} \right)\\x&=\sqrt{{{{{45}}^{2}}+{{{25}}^{2}}2\left( {45} \right)\left( {25} \right)\cos \left( {110} \right)}}\\&=58.48\end{align}\)
Now we can solve for the area of the two triangles, for the top triangle, it’s easier to use the formula with the angle: Top: \(\displaystyle \begin{align}\text{Area}&=\frac{1}{2}\left( a \right)\left( b \right)\left( {\sin C} \right)\\&=\frac{1}{2}\left( {45} \right)\left( {25} \right)\sin \left( {110} \right)\\&=528.58\end{align}\)
Bottom: \(\displaystyle \begin{align}s&=\frac{1}{2}\left( {a+b+c} \right)=86.74\\\text{Area}&=\sqrt{{86.74\left( {86.7455} \right)\left( {86.7460} \right)\left( {86.7458.48} \right)}}\\&=1442.38\end{align}\)
Adding the two areas together, we see that the total area of the piece of land is 1970.96 feet. 
Understand these problems, and practice, practice, practice!
Use the Right Triangle Button on the MathType keyboard to enter a problem, and then click on Submit (the arrow to the right of the problem) to solve the problem. You can also click on the 3 dots in the upper right hand corner to drill down for example problems.
If you click on “Tap to view steps”, you will go to the Mathway site, where you can register for the full version (steps included) of the software. You can even get math worksheets.
You can also go to the Mathway site here, where you can register, or just use the software for free without the detailed solutions. There is even a Mathway App for your mobile device. Enjoy!
On to Polar Coordinates, Equations and Graphs – you’re ready!