Review of Right Triangle Trig  Area of Triangles 
Law of Sines  Applications/Word Problems 
Law of Sines Ambiguous Case  More Practice 
Law of Cosines 
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; it’s unitless, since it’s basically a ratio.
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) for oblique triangles (triangles that don’t have a right angle). This can a little complicated, since we have to know which angles and sides we do have to know which of the “laws” to use. (Note that the Law of Sines can still be used to solve right triangles, using the 90° angle, which has a sin of 1, as one of the angles.)
Law of Sines
Again, we use the Law of Sines (or Sine Rule) to set up proportions to get other parts of a triangle that isn’t necessarily a right triangle.
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) (remember that these are “in a row” or adjacent parts of the triangle). 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; in these cases, use the Law of Cosines.
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, an acute (each angles measures less than 90°), or an obtuse (one of the angles measures more than 90°). For these cases, we have to account for both those shapes, so upon further study, we may have one triangle, two triangles with two sets of answers for each triangle, or no triangle. 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. Here is a summary for both Law of Sines and 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. 
Remember that if 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, instead of retyping 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.
Let’s do some problems; let’s first use the Law of Sines to find the indicated side or angle.
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: \(\displaystyle a=\frac{{\sin \left( {73} \right)\cdot 12}}{{\sin \left( {44} \right)}}\,\approx 16.52\) 
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: \(\displaystyle A={{\sin }^{{1}}}\left( {\frac{{\sin \left( {140} \right)\cdot 25}}{{45}}} \right)\approx20.9{}^\circ \) 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: \(\displaystyle b=\frac{{\sin \left( {74} \right)\cdot 42.2}}{{\sin \left( {43} \right)}}\approx59.48\) 
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  
\(m\angle A=25{}^\circ ,m\angle B=105{}^\circ ,a=450\)
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:
This makes sense, since larger angles have larger opposite sides! 
\(m\angle A=30{}^\circ ,m\angle B=110{}^\circ ,c=12.5\)
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:

\(a=22,\,\,\,b=14,\,\,\,\,m\angle A=110{}^\circ \)
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 \(\displaystyle \approx 36.7°\). Then we can get C: \(180(110+36.7) \approx 33.3°\). 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), since the triangle could either be acute or obtuse.
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 a sum of more than 180° for the triangle’s angles, there is no triangle that can be formed (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 then solve for two different triangles (the given two sides and one angle for the two triangles will be the same). If we 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 \(\displaystyle \sin A=\frac{h}{b}\). 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 \(b\) (see above): an obtuse with sides \(a\) and \({{c}_{1}}\), and an acute with sides \(a\) and \({{c}_{2}}\). Angles \({{B}_{1}}\) and \({{C}_{1}}\) will be associated with \(b\) and \({{c}_{1}}\), and angles \({{B}_{2}}\) and \({{C}_{2}}\) will be associated with \(b\) 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 six 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  
\(b=25,\,\,c=20\,,\,\,m\angle C=130{}^\circ \) 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 \(B\approx 73.2{}^\circ \). Then 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 32, for example (greater than \(b\), which is 25), we’d have one triangle and could solve the normal way using Law of Sines. 
\(b=14,\,\,\,c=20,\,\,\,\,m\angle B=40{}^\circ \) First draw the triangle (I like to draw the “paired” side to the right): 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 \(14<h\)). 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): \(C\approx66.7{}^\circ \). 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:

\(a=30,\,\,\,b=80,\,\,\,\,m\angle A=50{}^\circ \) 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 \(a<h\)).
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 \(h=b\sin A=80\sin \left( {50} \right)\approx61.3\), 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, \(6.427<a<10\) would result in two triangles. 
Law of Cosines
The Law of Cosines (or Cosine Rule) is a little bit more complicated, since it’s not a simple proportion.
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). (Remember that these are “in a row” or adjacent parts of the triangle). It also will work for the Side, Side, Angle (SSA), but the Law of Sines is usually taught with this case, because of the Ambiguous Case.
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. Here again is a summary for both Law of Sines and 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. 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; solve for \(a\): \(\displaystyle \begin{align}{{a}^{{2\,}}}&={{16}^{2}}+{{14}^{2}}2\left( {16} \right)\left( {14} \right)\cos \left( {100} \right)\\a&=\sqrt{{256+196\left( {448} \right)\left( {.173648} \right)}}\\&\approx 23.02\end{align}\)
(Note that angle A is called the included angle between sides b and c, since it is located between the two sides.)
Here’s how we can enter this problem in a graphing calculator (DEGREE mode): 
Solve for A: \(\displaystyle {{a}^{{2\,}}}={{b}^{2}}+{{c}^{2}}2bc\cos A\)
We have an SSS case; solve for A: \(\displaystyle \begin{align}{{14.5}^{2}}&={{10.5}^{2}}+{{12.2}^{2}}2\left( {10.5} \right)\left( {12.2} \right)\cos \left( A \right)\\{{14.5}^{2}}{{10.5}^{2}}{{12.2}^{2}}&=2\left( {10.5} \right)\left( {12.2} \right)\cos \left( A \right)\,\\\cos \left( A \right)&=\frac{{{{{14.5}}^{2}}{{{10.5}}^{2}}{{{12.2}}^{2}}}}{{2\left( {10.5} \right)\left( {12.2} \right)}}\\A&={{\cos }^{{1}}}\left( {\frac{{{{{14.5}}^{2}}{{{10.5}}^{2}}{{{12.2}}^{2}}}}{{2\left( {10.5} \right)\left( {12.2} \right)}}} \right)\approx 79.0{}^\circ \end{align}\)
Here’s how we can enter this problem in a graphing calculator (DEGREE mode); be sure to use the minus sign when subtracting in the numerator, and the negative sign (under the “3”) in the denominator: Or, in three steps, using “2^{nd} COS 2^{nd} () (ans)” 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: 
Solve for B:
\(\displaystyle {{b}^{{2\,}}}={{a}^{2}}+{{c}^{2}}2ac\cos B\)
We have an SSS case; 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: \(\require{cancel} 6+5\cancel{>}\,\,12\). 
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; solve for \(c\): \(\displaystyle \begin{align}{{c}^{{2\,}}}&={{3}^{2}}+{{5}^{2}}2\left( 3 \right)\left( 5 \right)\cos \left( {40} \right)\\c&=\sqrt{{9+25\left( {30} \right)\left( {.766044} \right)}}\\&\approx 3.32\end{align}\)
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:
\(\displaystyle \frac{{\sin A}}{3}=\frac{{\sin \left( {40} \right)}}{{3.32}};\,\,\,A\approx 35.5{}^\circ \)
Now we can subtract the sum of angles A and C from 180° to get angle B:

\(a=30,\,\,\,b=35,\,\,\,c=25\) \(\displaystyle {{a}^{{2\,}}}={{b}^{2}}+{{c}^{2}}2bc\cos A\)
We have an SSS case; solve for A first: \(\displaystyle \begin{align}{{30}^{{2\,}}}&={{35}^{2}}+{{25}^{2}}2\left( {35} \right)\left( {25} \right)\cos \left( A \right)\\A&={{\cos }^{{1}}}\left( {\frac{{{{{30}}^{2}}{{{35}}^{2}}{{{25}}^{2}}}}{{2\left( {35} \right)\left( {25} \right)}}} \right)\\&\approx 57.1{}^\circ \end{align}\)
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:
\(\displaystyle \frac{{\sin \left( {57.1} \right)}}{{30}}=\frac{{\sin C}}{{25}};\,\,\,C\approx 44.4{}^\circ \)
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 \(\angle C\):
\(\displaystyle \begin{align}{{c}^{{2\,}}}&={{a}^{2}}+{{b}^{2}}2ab\cos C\\{{12}^{2}}&={{8}^{2}}+{{10}^{2}}2\left( 8 \right)\left( {10} \right)\cos C\\C&={{\cos }^{{1}}}\left( {\frac{{{{{12}}^{2}}{{8}^{2}}{{{10}}^{2}}}}{{2\left( 8 \right)\left( {10} \right)}}} \right)\\C&={{\cos }^{{1}}}\left( {\frac{1}{8}} \right)\approx 82.8{}^\circ \end{align}\) 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  
\(a=14,\,\,\,b=20,\,\,\,\,m\angle A=40{}^\circ \) 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, we can also 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 get one value for c if one triangle can be formed, no value if no triangle can be formed, or two values if two triangles can be formed. In this case, two triangles can be formed since we get two different values (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: \(\displaystyle \frac{{\sin \left( {40} \right)}}{{14}}=\frac{{\sin C}}{{9.78}};\,\,\,C \approx 26.7{}^\circ \) \(\displaystyle \frac{{\sin 40}}{{14}}=\frac{{\sin C}}{{20.86}};\,\,\,C \approx 73.3{}^\circ \) 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: 1) two sides, and the angle inside the two sides (the SideAngleSide or SAS case), or 2) three sides of the triangle (SideSideSide, or SSS case). If we don’t initially have these cases, we can solve the triangle using the Law of Sines or Law of Cosines to get the angles and sides needed.
Here are the two formulas; note that one uses the sin function, and the other (Heron’s formula) uses the sides only:
Triangle  Area Formula  Use this Area Formula when you have these parts of a Triangle in a row: 
Area of Triangle: \(\displaystyle \begin{align}\text{Area}&=\frac{1}{2}\left( b \right)\left( c \right)\left( {\sin A} \right)\\\text{Area}&=\frac{1}{2}\left( a \right)\left( c \right)\left( {\sin B} \right)\\\text{Area}&=\frac{1}{2}\left( a \right)\left( b \right)\left( {\sin C} \right)\end{align}\) 

Area of Triangle (Heron’s): \(\displaystyle \text{Area}=\sqrt{{s\left( {sa} \right)\left( {sb} \right)\left( {sc} \right)}}\), where \(\displaystyle s=\,\,\frac{1}{2}\left( {a+b+c} \right)\,\,\,\,\,\text{(semiperimeter)}\) 
Here are some examples:
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; the area is straightforward: \(\displaystyle \begin{align}\text{Area}&=\frac{1}{2}\left( a \right)\left( b \right)\left( {\sin C} \right)\\&=\frac{1}{2}\left( 3 \right)\left( 5 \right)\sin \left( {40} \right)\approx 4.82\end{align}\) ———————————————————– Here’s one where we have to use Law of Sines first, since we don’t have the SAS case: \(b=100,\,\,\,c=75,\,\,\,m\angle C=30{}^\circ \) \(\displaystyle \frac{{\sin \left( {30} \right)}}{{75}}=\frac{{\sin \left( B \right)}}{{100}};\,\,\,\,\,B\approx41.8103{}^\circ ; \,\,\,A\approx 108.1897{}^\circ \) (extra digits for accuracy of next equation) \(\displaystyle \begin{align}\text{Area}&=\frac{1}{2}\left( b \right)\left( c \right)\left( {\sin A} \right)\\&=\frac{1}{2}\left( {100} \right)\left( {75} \right)\sin \left( {108.1897} \right)\approx3562.61\end{align}\) 
\(a=30,\,\,\,b=35,\,\,\,c=25\) \(\displaystyle \text{Area}=\sqrt{{s\left( {sa} \right)\left( {sb} \right)\left( {sc} \right)}}\), where \(\displaystyle s=\frac{1}{2}\left( {a+b+c} \right)\)
We have an SSS case; first find \(\displaystyle s:\,\,s=\frac{1}{2}\left( {30+35+25} \right)=45\). Then: \(\displaystyle \begin{align}\text{Area}&=\sqrt{{s\left( {sa} \right)\left( {sb} \right)\left( {sc} \right)}}\\&=\sqrt{{45\left( {4530} \right)\left( {4535} \right)\left( {4525} \right)}}\\&=\sqrt{{45\left( {15} \right)\left( {10} \right)\left( {20} \right)}}\\&\approx367.42\end{align}\)
Not too bad! Who figured out all this stuff? 
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, get the measurement of the third angle, which is 95°, and then use Law of Sines:
\(\displaystyle \frac{{\sin \left( {95} \right)}}{{250}}=\frac{{\sin \left( {45} \right)}}{b}\)
Cross multiply to get \(b\approx 177.45\).
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. 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 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). We know that \(AD=BC=450\); this is one of the properties of parallelograms: opposite sides are equal (congruent).
Using the Law of Cosines (SSS), get Angle B first: \(\displaystyle \begin{align}A{{C}^{2}}&=A{{B}^{2}}+B{{C}^{2}}2\left( {AB} \right)\left( {BC} \right)\cos \left( B \right)\\{{290}^{2}}&={{240}^{2}}+{{450}^{2}}2\left( {240} \right)\left( {450} \right)\cos \left( B \right)\\{{290}^{2}}{{240}^{2}}{{450}^{2}}&=2\left( {240} \right)\left( {450} \right)\cos \left( B \right)\\B&={{\cos }^{{1}}}\left( {\frac{{{{{290}}^{2}}{{{240}}^{2}}{{{450}}^{2}}}}{{2\left( {240} \right)\left( {450} \right)}}} \right)\\&\approx 35.4{}^\circ \end{align}\) Since two adjacent angles (two angles in a row) of a parallelogram add up to 180°, \(\angle BCD=180{}^\circ 35.4{}^\circ \approx 144.6{}^\circ \). 
Parallelogram \(ABCD\) has diagonals \(16\) and \(10\), with side \(\overline{{AD}}=7\).
What is the length of \(\overline{{CD}}\)?

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 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 \(y\)): \(\displaystyle \begin{align}A{{D}^{2}}&=D{{X}^{2}}+A{{X}^{2}}2\left( {DX} \right)\left( {AX} \right)\cos \left( y \right)\\{{7}^{2}}&={{8}^{2}}+{{5}^{2}}2\left( 8 \right)\left( 5 \right)\cos \left( y \right)\\{{7}^{2}}{{8}^{2}}{{5}^{2}}&=2\left( 8 \right)\left( 5 \right)\cos \left( y \right)\\y&={{\cos }^{{1}}}\left( {\frac{{{{7}^{2}}{{8}^{2}}{{5}^{2}}}}{{2\left( 8 \right)\left( 5 \right)}}} \right)\\&=60{}^\circ \end{align}\) Use the Law of Cosines again to get the length of \(\overline{{CD}}\), 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 \(180{}^\circ \)): \(\displaystyle \begin{align}C{{D}^{2}}&=D{{X}^{2}}+X{{C}^{2}}2\left( {DX} \right)\left( {XC} \right)\cos \left( {180y} \right)\\C{{D}^{2}}&={{8}^{2}}+{{5}^{2}}2\left( 8 \right)\left( 5 \right)\cos \left( {120} \right)\\C{{D}^{2}}&=129\\CD&\approx 11.36\end{align}\)$ 
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.
Bearing Problem
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, like when we read a map, 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 or SW, for example, the angle would be 45° south of west, or 225° clockwise from north, and so on.
Remember that each time an object changes course, you have to draw another line to the north to map its new bearing. Here are some bearing examples:
Here is a 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. (Note that this problem is actually solved more easily using vectors here in the Introduction to Vectors section.)
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 (bearing from ending to starting 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 roughly 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 roughly \(40°+35.6°= 75.6°\) (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 (try to draw as proportionate as possible): 
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).
Because all the angles in a right angle (perfect corner) add up to 90°, we can see that the angle inside the triangle near Joa is \(90°–(25+30)°=35°\).
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: \(\displaystyle \begin{align}\frac{{\sin \left( {110} \right)}}{{100}}&=\frac{{\sin \left( {35} \right)}}{x}\\x&\approx 61.039\end{align}\) 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. 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 make two triangles with the figure, and then start filling in the “pieces of the puzzle” with what we can solve. Call the diagonal “\(x\)” as shown; we’ll need this to get the area of the lower triangle.
Solve for \(x\), using the top triangle and the Law of Cosines (SAS): \(\displaystyle \begin{align}{{x}^{2}}&={{45}^{2}}+{{25}^{2}}2\left( {45} \right)\left( {25} \right)\cos \left( {110} \right)\\x&=\sqrt{{{{{45}}^{2}}+{{{25}}^{2}}2\left( {45} \right)\left( {25} \right)\cos \left( {110} \right)}}\\&\approx 58.48\end{align}\)
Now solve for the area of the two triangles; for the top triangle, it’s easier to use the formula with the angle: Top: \(\displaystyle \begin{align}\text{Area}&=\frac{1}{2}\left( a \right)\left( b \right)\left( {\sin C} \right)\\&=\frac{1}{2}\left( {45} \right)\left( {25} \right)\sin \left( {110} \right)\\&\approx 528.58\end{align}\)
Bottom: \(\displaystyle \begin{align}s&=\frac{1}{2}\left( {a+b+c} \right)\approx 86.74\\\text{Area}&=\sqrt{{s\left( {sa} \right)\left( {sb} \right)\left( {sc} \right)}}\\\text{Area}&=\sqrt{{86.74\left( {86.7455} \right)\left( {86.7460} \right)\left( {86.7458.48} \right)}}\\&\approx 1442.38\end{align}\)
Add the two areas together; the total area of the piece of land is roughly 1970.96 feet. 
Understand these problems, and practice, practice, practice!
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On to Polar Coordinates, Equations and Graphs – you’re ready!