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You may have been introduced to Trigonometry in Geometry, when you had to find either a side length or angle measurement of a triangle. Trigonometry is basically the study of triangles, and was first used to help in the computations of astronomy. Today it is used in engineering, architecture, medicine, physics, among other disciplines.
The 6 basic trigonometric functions that you’ll be working with are sine (rhymes with “sign”), cosine, tangent, cosecant, secant, and cotangent. (Don’t let the fancy names scare you; they really aren’t that bad).
With Right Triangle Trigonometry, we use the trig functions on angles, and get a number back that we can use to get a side measurement, as an example. Sometimes we have to work backwards to get the angle measurement back so we have to use what a call an inverse trig function. But basically remember that we need the trig functions so we can determine the sides and angles of a triangle that we don’t otherwise know.
Later, we’ll see how to use trig to find areas of triangles, too, among other things.
You may have been taught SOH – CAH – TOA (SOHCAHTOA) (pronounced “so – kuh – toe – uh”) to remember these. Back in the old days when I was in high school, we didn’t have SOHCAHTOA, nor did we have fancy calculators to get the values; we had to look up trigonometric values in tables.
Remember that the definitions below assume that the triangles are right triangles, meaning that they all have one right angle (90°). Also note that in the following examples, our angle measurements are in degrees; later we’ll learn about another angle measurement unit, radians, which we’ll discuss here in the Angles and Unit Circle section.
Basic Trigonometric Functions (SOH – CAH – TOA)
Here are the 6 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! Note that the cosecant (csc), secant (sec), and cotangent (cot) functions are called reciprocal functions, or reciprocal trig functions, since they are the reciprocals of sin, cos, and tan, respectively.
For the coordinate system method, assume that the vertex of the angle in the triangle is at the origin \((0,0)\):
Right Triangle 
SOHCAHTOA Method 
Coordinate System Method 
\(\displaystyle \begin{align}\text{SOH: Sine}\left( A \right)=\sin \left( A \right)=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}\\\text{CAH: Cosine}\left( A \right)=\cos \left( A \right)=\frac{{\text{Adjacent}}}{{\text{Hypotenuse}}}\\\text{TOA: Tangent}\left( A \right)=\tan \left( A \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}\end{align}\)
\(\displaystyle \begin{align}\text{cosecant}\left( A \right)=\csc \left( A \right)=\frac{1}{{\sin \left( A \right)}}=\frac{{\text{ Hypotenuse}}}{{\text{Opposite}}}\\\text{secant}\left( A \right)=\sec \left( A \right)=\frac{1}{{\cos \left( A \right)}}=\frac{{\text{ Hypotenuse}}}{{\text{Adjacent}}}\\\text{cotangent}\left( A \right)=\cot \left( A \right)=\frac{1}{{\tan \left( A \right)}}=\frac{{\text{ Adjacent}}}{{\text{Opposite}}}\end{align}\) 
\(\displaystyle \begin{align}\sin \left( A \right)=\frac{y}{h}\\\cos \left( A \right)=\frac{x}{h}\\\tan \left( A \right)=\frac{y}{x}\end{align}\)
\(\displaystyle \begin{align}\csc \left( A \right)=\frac{1}{{\sin \left( A \right)}}=\frac{h}{y}\\\sec \left( A \right)=\frac{1}{{\cos \left( A \right)}}=\frac{h}{x}\\\cot \left( A \right)=\frac{1}{{\tan \left( A \right)}}=\frac{x}{y}\end{align}\) 
Here are some example problems. Note that we commonly use capital letters to represent angle measurements, and the same letters in lower case to represent the side measurements opposite those angles.
We also use the theta symbol θ to represent angle measurements, as we’ll see later. Note also in these problems, we need to put our calculator in the DEGREE mode.
And don’t forget the Pythagorean Theorem (\({{a}^{2}}+{{b}^{2}}={{c}^{2}}\), where \(a\) and \(b\) are the “legs” of the triangle, and \(c\) is the hypotenuse), and the fact that the sum of all angles in a triangle is 180°.
Right Triangle Problem  Explanation  Calculator Steps/Checking 
Find the values of \(a\) and \(b\):
Once we get the answers, we can check our sides using the Pythagorean Theorem: \(\begin{array}{c}{{a}^{2}}+{{b}^{2}}={{c}^{2}}\\{{\left( {16.383} \right)}^{2}}+{{\left( {11.471} \right)}^{2}}=399.99\\\approx {{\left( {20} \right)}^{2}}\end{array}\) 
To get side \(a\), we need to use:
\(\displaystyle \sin \left( A \right)=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}\), where \(A\) is 55°: \(\displaystyle \sin \left( {55{}^\circ } \right)=\frac{a}{{20}}\) Cross multiply: \(\displaystyle a=\sin \left( {55{}^\circ } \right)\cdot 20\approx 16.383\)
To get side \(b\), we need to use: \(\displaystyle \cos \left( A \right)=\frac{{\text{Adjacent}}}{{\text{Hypotenuse}}}\), where \(A\) is 55°: \(\displaystyle \cos \left( {55{}^\circ } \right)=\frac{b}{{20}}\) Cross multiply: \(\displaystyle b=\cos \left( {55{}^\circ } \right)\cdot 20\approx 11.472\) 
Hit and scroll down and to the right to make sure you’re in DEGREE mode.
Then use theandkeys for cosine and sine, respectively: 
Find the value of \(b\):  To get side \(b\), we need to use:
\(\displaystyle \tan \left( A \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}\), where \(A\) is 23°: \(\displaystyle \tan \left( {23{}^\circ } \right)=\frac{6}{b}\) Cross multiply: \(\displaystyle \tan \left( {23{}^\circ } \right)\cdot b=6\), or turn proportion sideways with an \(“=”\) sign: \(\displaystyle \frac{b}{1}=\frac{6}{{\tan \left( {23{}^\circ } \right)}};\,\,\,\,a\approx 14.135\) 
Use the key:
If we needed to also find \(h\), we could either use \(\displaystyle \sin \left( {23{}^\circ } \right)=\frac{6}{h}\) or Pythagorean Theorem; both ways reveal that \(h=15.356\). 
Find the values of \(A\) and \(B\):
Once we get all the answers, let’s check to make sure the sum of all angles is 180°: \(\displaystyle \begin{array}{c}51.1{}^\circ \text{ }+38.9{}^\circ +90{}^\circ \left( {\text{right angle}} \right)\\=180{}^\circ \end{array}\) 
This one’s a little trickier since we need to find angle measurements instead of side measurements; we’ll need to use the \({{\sin }^{{1}}}\left( A \right)\) and \({{\cos }^{{1}}}\left( A \right)\) (2^{nd} sin and 2^{nd} cos on the calculator) to get the angles back.
For angle \(A\), we can use sin, since we have the opposite side (14) and hypotenuse (18): \(\displaystyle \begin{align}\sin \left( A \right)&=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}=\frac{{14}}{{18}}\\A&={{\sin }^{{1}}}\left( {\frac{{14}}{{18}}} \right)\approx 51.1{}^\circ \end{align}\)
For angle \(B\), we use cos: \(\displaystyle \begin{align}\cos \left( B \right)&=\frac{{\text{Adjacent}}}{{\text{Hypotenuse}}}=\frac{{14}}{{18}}\\B&={{\cos }^{{1}}}\left( {\frac{{14}}{{18}}} \right)\approx 38.9{}^\circ \end{align}\) 
Use theand keys: 
Find the values of \(x\) and \(y\):

This is a difficult problem that can easily be solved using Law of Sines, but let’s solve it by drawing an altitude to the triangle (dotted line). This creates two right triangles.
To get altitude \(a\), we can use: \(\displaystyle \sin \left( A \right)=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}\), where \(A\) is 25°: \(\displaystyle \sin \left( {25{}^\circ } \right)=\frac{a}{{35}};\,\,\,a=\sin \left( {25{}^\circ } \right)\cdot 35;\,\,\,\,a\approx 14.7916\)
Now we can use altitude \(a\) to get side \(x\), using the second right triangle: \(\displaystyle \sin \left( A \right)=\frac{{\text{Opposite}}}{{\text{Hypotenuse}}}\), where \(A\) is 20°: \(\displaystyle \sin \left( {20{}^\circ } \right)=\frac{{14.7916}}{x};\,\,\,x=\frac{{14.7916}}{{\sin \left( {20{}^\circ } \right)}};\,\,\,\,x\approx 43.2477\)
Now we can use the Pythagorean Theorem to get the two parts of bottom \(y\) (we could have also used right angle trig): \(\displaystyle \begin{array}{c}{{y}_{1}}^{2}+{{a}^{2}}={{35}^{2}};\,\,\,\,\,{{y}_{1}}^{2}+{{14.7916}^{2}}={{35}^{2}};\,\,\,\,{{y}_{1}}\approx 31.7208\\{{y}_{2}}^{2}+{{a}^{2}}={{x}^{2}};\,\,\,\,\,{{y}_{2}}^{2}+{{14.7916}^{2}}={{43.2477}^{2}};\,\,\,\,{{y}_{2}}\approx 40.6395\\y={{y}_{1}}+{{y}_{2}}\approx 72.3603\end{array}\) Whew! Hard problem! 
Here are some problems where we need to think about ratios of sides of right triangles:
Right Triangle Ratio Problem  Math 
In the right triangle ABC below, if \(\displaystyle \frac{{BC}}{{AB}}=\frac{2}{5}\), the exact value of \(\tan \left( A \right)=?\).  Since \(\displaystyle \frac{{BC}}{{AB}}=\frac{2}{5}\), from the perspective of \(\angle A\), we have information on the opposite and hypotenuse sides.
To get \(\tan \left( A \right)\), we need to also get the adjacent side to A. We can use the Pythagorean Theorem to get this, since the side ratios will still be intact, regardless of the size of the triangle. Let’s just “pretend” that \(BC=2\) and \(AB=5\): \(B{{C}^{2}}+A{{C}^{2}}=A{{B}^{2}};\,\,\,\,{{2}^{2}}+A{{C}^{2}}={{5}^{2}};\,\,\,AC=\sqrt{{{{5}^{2}}{{2}^{2}}}}=\sqrt{{21}}\) Thus, \(\displaystyle \tan \left( A \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}=\frac{{BC}}{{AC}}=\frac{2}{{\sqrt{{21}}}}\,\,\,(=\frac{2}{{\sqrt{{21}}}}\cdot \frac{{\sqrt{{21}}}}{{\sqrt{{21}}}}=\frac{{2\sqrt{{21}}}}{{21}})\) Try it – it works! 
Given triangle \(EDF\) with a right angle at \(F\), side \(d=9\) and side \(f=12\).
The exact value of \(\csc \left( E \right)=?\) 
Draw a picture first:
To get \(\csc \left( E \right)\), we need to also get the opposite side of \(E\). We can use the Pythagorean Theorem to get this: \({{d}^{2}}+{{e}^{2}}={{f}^{2}};\,\,\,\,{{9}^{2}}+{{e}^{2}}={{12}^{2}};\,\,\,e=\sqrt{{{{{12}}^{2}}{{9}^{2}}}}=\sqrt{{63}}=3\sqrt{7}\) Thus, \(\displaystyle \csc \left( E \right)=\frac{{\text{Hypotenuse}}}{{\text{Opposite}}}=\frac{{12}}{{3\sqrt{7}}}=\frac{4}{{\sqrt{7}}}\,\,\,(=\frac{4}{{\sqrt{7}}}\cdot \frac{{\sqrt{7}}}{{\sqrt{7}}}=\frac{{4\sqrt{7}}}{7})\) 
Trigonometry Word Problems
Here are some types of word problems (applications) that you might see when studying right angle trigonometry.
Note that the angle of elevation is the angle up from the ground; for example, if you look up at something, this angle is the angle between the ground and your line of site.
The angle of depression is the angle that comes down from a straight horizontal line in the sky. (For example, if you look down on something, this angle is the angle between your looking straight and your looking down to the ground). For the angle of depression, you can typically use the fact that alternate interior angles of parallel lines are congruent (sorry, too much Geometry!) to put that angle in the triangle on the ground (we’ll see examples). Note that shadows in these types of problems are typically on the ground. When the sun casts the shadow, the angle of depression is the same as the angle of elevation from the ground up to the top of the object whose shadow is on the ground.
Also, the grade of something (like a road) is the tangent (rise over run) of that angle coming from the ground. Usually the grade is expressed as a percentage, and you’ll have to convert the percentage to a decimal to use in the problem.
And, as always, always draw pictures!
Angle of Elevation Problems:
Angle of Elevation Trig Problem  Math 
Devon is standing 100 feet from the Eiffel Tower and sees a bird land on the top of the tower (she has really good eyes!).
If the angle of elevation from Devon to the top of the Eiffel Tower is close to 84.6°, how tall is the tower? 
This is a good example how we might use trig to get distances that are typically difficult to measure. Note that the angle of elevation comes up off of the ground.
To get the height \(y\), we need to use: \(\displaystyle \tan \left( \theta \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}\), where \(\theta =84.6{}^\circ \) \(\displaystyle \tan \left( {84.6{}^\circ } \right)=\frac{y}{{100}}\) Cross multiply to get: \(\displaystyle y=\tan \left( {84.6{}^\circ } \right)\cdot 100\,\,\approx \,\,1058\).
The Eiffel Tower is roughly 1058 feet tall. 
A surveyor stands 100 meters from the bottom of a building, and measures the angle (of elevation) to the top of the building to be 50°. She then measures the angle (of elevation) to the top of the tower on the roof to be 60°.
How tall is the tower on the roof?

The first step is to draw a picture, and assign variables to what we want, using what we have. Let \(x\) equal the height of the tower, and \(y\) the height of the building.
Use two tangent functions, first obtaining \(y\):
To get \(y\): \(\displaystyle \tan \left( {50{}^\circ } \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}=\frac{y}{{100}};\,\,\,y=\tan \left( {50{}^\circ } \right)\times 100\approx 119.18\) To get \(x+y\): \(\displaystyle \tan \left( {60{}^\circ } \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}=\frac{{x+y}}{{100}};\,\,\,\,x+y=\tan \left( {60{}^\circ } \right)\times 100\approx 173.21\)
To get \(x\), we subtract \(y\) from \(x+y\), so the tower is \(173.21119.18\approx 54.03\) meters. 
Angle of Depression Problem:
Right Triangle Systems Problem:
Here’s a problem where it’s easiest to solve it using a System of Equations:
Right Triangle Systems Trig Problem  Math 
Two girls are standing 100 feet apart. They both see a beautiful seagull in the air between them. The angles of elevation from the girls to the bird are 20° and 45°, respectively.
How high up is the seagull? 
The trick here is divide up the 100 ft into \(x\) and the other is \(100x\) (put in real numbers to see how we get this) since we have two triangles. Then we will have two equations (one for each triangle) and two unknowns:
\(\displaystyle \tan \left( {20{}^\circ } \right)=\frac{y}{x};\,\,\,\,\,\,\,\,\,\,\,\,\tan \left( {45{}^\circ } \right)=\frac{y}{{100x}}\)
From the first equation, we get \(y\) in terms of \(x\): \(\displaystyle y=\tan \left( {20{}^\circ } \right)\cdot x\approx .36397x\). We can plug this into the second equation to get \(\displaystyle \tan \left( {45{}^\circ } \right)=\frac{{.36397x}}{{100x}}\). Solving for \(x\), we get: \(\displaystyle 1=\frac{{.36397x}}{{100x}};\,\,\,\,100x=.36397x;\,\,\,\,\,x\approx 73.315\). (I left more decimal places, so the final answer will be more accurate).
Now we have to get \(y\) to find the height of the seagull: \(\displaystyle y=.36397\left( {73.315} \right)=27.684\)
The seagull is about 26.68 feet high. This was a tricky one! 
Trig Shadow Problem:
Trig Shadow Problem  Math 
The length of a tree’s shadow is 20 feet when the angle of elevation to the sun is 40°. How tall is the tree?  Again, note that shadows in these types of problems are on the ground. When the sun casts the shadow, the angle of depression is the same as the angle of elevation from the ground up to the top of the tree.
To get the height \(y\) , we need to use: \(\displaystyle \tan \left( \theta \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}\), where \(\theta =40{}^\circ \) \(\displaystyle \tan \left( {40{}^\circ } \right)=\frac{x}{{20}}\)
Cross multiply to get \(\displaystyle x=\tan \left( {40{}^\circ } \right)\cdot 20\approx 16.78\).
The height of the tree is approximately 16.78 feet tall. Not too bad! 
Trig Grade Problem:
Trig Grade Problem  Math 
Chelsea walked up a road that has a 20% grade (she could feel it!) to get to her favorite store.
At what angle does the road come up from the ground (at what angle is the road inclined from the ground)?

Remember that the grade of a road can be thought of as \(\displaystyle \frac{{\text{rise}}}{{\text{run}}}\), and you usually see it as a percentage. So, a 20% grade is the same as a grade of \(\displaystyle \frac{{\text{20}}}{{\text{100}}};\) for every 20 feet the road goes up vertically, it goes 100 feet horizontally.
Since we need to find an angle measurement and we have the adjacent and opposite sides, we’ll need to use the \({{\tan }^{{1}}}\left( \theta \right)\) (2^{nd} tan on the calculator and make sure it’s in DEGREE mode) to get the angle back: \(\displaystyle \tan \left( \theta \right)=\frac{{\text{Opposite}}}{{\text{Adjacent}}}=\frac{{20}}{{100}};\,\,\,\,\,\,\,\,\theta ={{\tan }^{{1}}}\left( {\frac{{20}}{{100}}} \right)=11.31{}^\circ \) The road comes up at an angle of roughly 11.31° from the ground.
Note that if we wanted to know how long the actual slanted road is, we could just use Pythagorean Theorem, or sin or cos: \(\displaystyle \sin \left( {11.31{}^\circ } \right)=\frac{{20}}{x};\,\,\,\,x=\frac{{20}}{{\sin \left( {11.31{}^\circ } \right)}}\approx 101.98\text{ }ft\) This makes sense since the grade is relatively small (note that the picture is not drawn to scale!)

Understand these problems, and practice, practice, practice!
Click on Submit (the arrow to the right of the problem) to solve this problem. You can also type in more problems, or click on the 3 dots in the upper right hand corner to drill down for example problems.
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On to Angles and the Unit Circle – you’re ready!