Topic outline
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By the end of this course you will be able to:
- Define and use the trigonometric ratios of \(cos\theta \), \(sin\theta \), and \(tan\theta \).
- Calculate the value of expressions containing trigonometric ratios with a calculator.
- Calculate the value of expressions containing trigonometric ratios without a calculator.
- Work with trigonometric tables.
- Solve problems in 2 dimensions involving angles of elevation and depression
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Before you start this course, you need to make sure that you know and can use the Theorem of Pythagoras to find the length of unknowns sides in right angled triangles.
The Theorem of Pythagoras states that for any right-angled triangle the square of the length of the hypotenuse (always the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In other words, in the triangle below, it means that \( c^2=a^2+b^2 \).
Try answering the following question to make sure you understand the theorem and how to use it.
In \( \Delta ABC \), \( \hat {A}=90 ^\circ \), \(\text{AB}=4.5\,\text{cm}\) and \(\text{BC}=7\,\text{cm}\). Calculate the length of \(\text{AC}\).
Solution:
\( \Delta \text{ABC}\) is a right-angled triangle. Therefore, we can use the Theorem of Pythagoras to calculate the length of the \( \text{AC}\).
\( \begin{align}\text{B}{{\text{C}}^{2}}&=\text{A}{{\text{B}}^{2}}+\text{A}{{\text{C}}^{2}}\\\therefore \text{A}{{\text{C}}^{2}} &=\text{B}{{\text{C}}^{2}}-\text{A}{{\text{B}}^{2}}\\\therefore \text{AC}&=\sqrt{{\text{B}{{\text{C}}^{2}}-\text{A}{{\text{B}}^{2}}}}\\ &=\sqrt{{{{7}^{2}}-{{{4.5}}^{2}}}}\\ &=\sqrt{{49-20.25}}\\&=\sqrt{{28.75}}\\&=5.36\end{align} \)
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What is trigonometry (or ‘trig’ as it is sometimes called)? A clue lies in the first part of the word ‘trigonometry’. ‘Tri’ means three; as in tricycle (three wheels) and triangle (three angles and three sides). Trigonometry deals with the relationships between the angles and sides of triangles.
The word trigonometry comes from the Greek words for triangle (trigōnon) and measure (metron).
From about the year 150 A.D., Egyptians and Greeks used trigonometry to measure the distances between objects and places they could not measure directly. They even used it to measure the distances between stars and the circumference of the Earth. Trigonometry uses a technique called triangulation to measure distances.
Trigonometry has, therefore, always played a key part in navigation. Many of the modern applications of trigonometry still have to do with navigation. Modern satellite navigation systems still use trigonometry and triangulation to find the distances between landmarks.
Artist’s concept of a NAVSTAR Global Positioning System satellite, a space-based radio navigation network
Other fields that use trigonometry include acoustics, optics, financial market analysis, electronics, probability, statistics, biology, medical imaging, chemistry, cryptology, meteorology, oceanography, land surveying, architecture, phonetics, engineering, computer graphics, and game development, amongst many others. -
To lay the foundation for understanding what trigonometry is and how it works, do this next activity.
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Activity 1: Investigate the sides of similar triangles
Time required: 20 minutes
What you need:
- a pen or pencil
- paper
- a protractor
- a ruler
What to do:
- Draw three triangles of different sizes using a protractor and a ruler so that each triangle has interior angles equal to \( 30^ \circ \),0\( 60^ \circ \) and \( 90^ \circ \) as shown below. We call these ‘similar triangles’ because they are all the same shape, but not the same size. If you do not have a protractor or ruler, you can measure the lengths of the triangles below instead of your own.
- Now measure the angles and lengths accurately and complete the following table. Leave all your ratios as fractions for now.
- What do you notice about each of the ratios? You can convert each ratio to a decimal if this helps you to compare. Does it matter what the lengths of the sides of the triangles are if the angles inside the three triangles stay the same size?
- Draw one more right-angled triangle but this time make the other two angles \( 40^ \circ \) and \( 50^ \circ \). Measure the lengths of the sides as before. Are the sides in the same ratio as the first three triangles that you drew?
What did you find?
You should have found that the value of each of the ratios of corresponding sides in each triangle was always the same. In other words, \( \displaystyle \frac{{\text{AB}}}{{\text{BC}}}=\displaystyle \frac{{\text{DE}}}{{\text{EF}}}=\displaystyle \frac{{\text{GH}}}{{\text{HK}}}\), \( \displaystyle \frac{{\text{AB}}}{{\text{AC}}}=\displaystyle \frac{{\text{DE}}}{{\text{DF}}}=\displaystyle \frac{{\text{GH}}}{{\text{GK}}}\) and \( \displaystyle \frac{{\text{BC}}}{{\text{AC}}}=\displaystyle \frac{{\text{EF}}}{{\text{DF}}}=\displaystyle \frac{{\text{HK}}}{{\text{GK}}}\).This means that it is the size of the angles in a triangle that determines the ratios between the lengths of the sides. It does not matter how big or small the triangle is. If the sizes of the angles stay the same, the corresponding sides will always be in the same ratio. If we change the sizes of the angles, we change the ratio of the sides.
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Activity 2: Optional internet activity
If you have an internet connection, continue to explore how it is the angles inside a right-angled triangle that determine the ratio of the sides of the triangle by playing with the Right Triangle Ratios interactive tool. Drag the slider to change the angle. See how the ratios of the sides change? Drag point A left and right to change the size of the triangle with a given angle. See how the ratio of the sides stays the same?
What ratios do you get when you make the angle \( 30^ \circ \) or \( 60^ \circ \).
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We use the fact that it is the size of the angles in a triangle that determine the ratio of the sides as the basis of trigonometry. We can define three basic trigonometric ratios for any triangle.
Have a look at the right-angled triangle below. We can name the sides of the triangle with reference to the angle called \( \theta \) (theta) at as follows. The hypotenuse () is always the side opposite the right-angle. The side next to () is called the adjacent side (adjacent means 'next to') and the side opposite \( \theta \) () is called the opposite side.
If we defined the sides with respect to \( \beta \) (beta) at , then the adjacent side would be and the opposite side would be . The hypotenuse would remain unchanged.
Note The definitions of opposite, adjacent and hypotenuse are only applicable when working with right-angled triangles. Always check to make sure your triangle has a right-angle before you use them.
We can define the three ratios of the lengths of the sides within any right-angled triangle. We can give each of these three ratios a special name – sine, cosine and tangent – all with respect to the angle \( \theta \):
\( \sin \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the hypotenuse}} \)
\(\cos \theta=\displaystyle \frac{\text{length of the adjacent side}}{\text{length of the hypotenuse}} \)
\(\tan \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the adjacent side}} \)
With respect to angle \( \theta \) in \( \Delta ABC \) above, these three ratios would be:
\( \sin \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the hypotenuse}}=\frac{BC}{AB}\)
\(\cos \theta=\displaystyle \frac{\text{length of the adjacent side}}{\text{length of the hypotenuse}}=\frac{AC}{AB} \)
\(\tan \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the adjacent side}}=\frac{BC}{AC} \)
Can you work out what these three ratios would be when defined with respect to angle \( \beta )\. Write them down.
With respect to angle \( \beta \) in \( \Delta ABC \) above, these three ratios would be:
\( \sin \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the hypotenuse}}=\displaystyle \frac{AC}{AB}\)
\(\cos \theta=\displaystyle \frac{\text{length of the adjacent side}}{\text{length of the hypotenuse}}=\displaystyle \frac{BC}{AB} \)
\(\tan \theta=\displaystyle \frac{\text{length of the opposite side}}{\text{length of the adjacent side}}=\displaystyle \frac{AC}{BC} \)
The three ratios of sine, cosine and tangent form the basis of all of trigonometry.
Note The three basic trigonometric ratios are:
\( \sin \theta=\displaystyle \frac{\text{opposite}}{\text{hypotenuse}} \)
\(\cos \theta=\displaystyle \frac{\text{adjacent}}{\text{hypotenuse}} \)
\(\tan \theta=\displaystyle \frac{\text{opposite}}{\text{adjacent}} \)
We never define the trigonometric ratios with respect to the right-angle.
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Activity 1: Example 1
State the three basic trigonometric ratios for the following triangles with respect to the angle \( \theta \).
- \( \sin \theta=\displaystyle \frac{\text{opp}}{\text{hyp}} =\displaystyle\frac{EF}{DE} \)
- \( \cos \theta=\displaystyle \frac{\text{adj}}{\text{hyp}} =\displaystyle\frac{DF}{DE} \)
- \( \tan \theta=\displaystyle \frac{\text{opp}}{\text{adj}} =\displaystyle\frac{EF}{DF} \)
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Activity 2: Example 2
State each of the following ratios for the given triangle:
- \( \sin\hat{A} \)
- \( \cos\hat{A} \)
- \( \tan\hat{A} \)
- \( \sin\hat{C} \)
- \( \cos\hat{C} \)
- \( \tan\hat{C} \)
Solution:
- \( \sin\hat{A}=\displaystyle\frac{\text{opp}}{\text{hyp}}=\displaystyle\frac{BC}{AC} \)
- \( \cos\hat{A}=\displaystyle\frac{\text{adj}}{\text{hyp}}=\displaystyle\frac{AB}{AC} \)
- \( \tan\hat{A}=\displaystyle\frac{\text{opp}}{\text{adj}}=\displaystyle\frac{BC}{AB} \)
- \( \sin\hat{C}=\displaystyle\frac{\text{opp}}{\text{hyp}}=\displaystyle\frac{AB}{AC} \)
- \( \cos\hat{C}=\displaystyle\frac{\text{opp}}{\text{hyp}}=\displaystyle\frac{BC}{AC} \)
- \( \tan\hat{C}=\displaystyle\frac{\text{opp}}{\text{hyp}}=\displaystyle\frac{AB}{BC} \)
- \( \sin\hat{A} \)
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SOH CAH TOA
You can use Soh Cah Toa to help you remember how each of the trig ratios are defined.
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Activity 3: Example 3
Have a look at this next example.
Given \( \Delta ABC \) with \( \hat{C}=90^\circ \),\( \hat{A}=\theta \) and\( \hat{B}=\alpha \).
Determine the value of the following, showing all calculations.
- \( \cos\alpha \)
- \( \tan\theta \)
- \( \displaystyle\frac{\sin\alpha}{\cos\alpha} \)
- \( 1+\tan^2\alpha \)
Solution
- \( \cos\alpha=\displaystyle\frac{\text{adj}}{\text{hyp}}=\displaystyle\frac{BC}{AB}=\displaystyle\frac{4}{5} \)
- \( \tan\theta=\displaystyle\frac{\text{opp}}{\text{adj}}=\displaystyle\frac{BC}{AC} \).
But we do not know the length of AC. However, we can use the theorem of Pythagoras to find AC.
\( \begin{align*}AB^2&=AC^2+BC^2 \\\therefore AC^2 &= AB^2-BC^2 \\\therefore AC&=\sqrt{AB^2-BC^2} \\&=\sqrt{5^2-4^2} \\&=\sqrt{25-16} \\&=\sqrt{9} \\&=3 \end{align*} \)
Therefore \( \tan\theta=\displaystyle\frac{\text{opp}}{\text{adj}}=\displaystyle\frac{BC}{AC}=\displaystyle\frac{4}{3} \) - \( \displaystyle\frac{\sin\alpha}{\cos\alpha}=\displaystyle\frac{\frac{3}{5}}{\frac{4}{5}}=\displaystyle\frac{3}{5} \times\displaystyle\frac{5}{4}=\displaystyle\frac{3}{4} \)
- Remember that \( \tan\theta \) is actually just a number (a fraction). Therefore, we can apply all the normal operations to the trigonometric ratios. In this case we have to square the ratio and add 1 to it.
\( 1+\tan^2\theta=1+(\tan\theta)^2=1+ \left( \displaystyle\frac{4}{3}\right)^2=1+\displaystyle\frac{16}{9} = \displaystyle\frac{9}{9}+\displaystyle\frac{16}{9}=\displaystyle\frac{25}{9} \)
Note If we mean to raise a trigonometric ratio to a power, we always write the power directly after the ratio like \( \sin^2\theta \) or \( \cos^3\alpha \). If you see \( \sin\theta^2 \), for example, this means that the angle is being raised to the power not the whole ratio. - \( \cos\alpha \)
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Almost always, you will have to work with triangles where the angles are not \( 30^\circ \), \( 45^\circ \), or \( 60^\circ \). In these cases, you will need a scientific calculator to help you. However, when the triangle you are working with does contain \( 30^\circ \), \( 45^\circ \), or \( 60^\circ \), you should recognise these as the special angles and remember what the corresponding trigonometric ratios are.
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Activity 1: Using a calculator
Question: Use a calculator to work out what \( \tan 34.16^\circ \) is.
Answer:
If your calculator is direct input (e.g. most Casio calculators) key in the question like this:
\( \boxed{\tan}\to \boxed{34.16}\to\boxed{)}\to\boxed{=} \)
If you calculator is not direct input (e.g. most Sharp calculators_ key in the question like this:
\( \boxed{34.16}\to\boxed{\tan}\to\boxed{=} \)
In either case, your calculator should give you that \( \tan 34.16{}^\circ =0.678579...\approx 0.679 \). -
Activity 2: The special angles
Question: Have a look at the three right-angled triangles below.
Now draw a table like this and work out the various trig ratios to complete it.
\( 30^\circ \) \( 45^\circ \) \( 60^\circ \) \( \sin \) \( \cos \) \( \tan \) Answer: You should find that your table looks like this
\( 30^\circ \) \( 45^\circ \) \( 60^\circ \) \( \sin \) \( \displaystyle\frac{1}{2} \) \( \displaystyle\frac{1}{\sqrt{2}} \) \( \displaystyle\frac{\sqrt{3}}{2} \) \( \cos \) \( \displaystyle\frac{\sqrt{3}}{2} \) \( \displaystyle\frac{1}{\sqrt{2}} \) \( \displaystyle\frac{1}{2} \) \( \tan \) \( \displaystyle\frac{1}{\sqrt{3}} \) \( \displaystyle\frac{1}{1} \) \( \displaystyle\frac{\sqrt{3}}{1} \)
\( 30^\circ \), \( 45^\circ \) and \( 60^\circ \) are called special angles because we don't need to use a calculator to work them out. They give ratios that are easy to right down as fractions and these fractions express their value exactly.
Remember when we calculated the value of \( \tan 34.16^\circ \) above? The calculator gave us the answer \( =0.678579... \). The calculator had to round the answer off. We then rounded the answer off further to \( \approx 0.679 \). Therefore, the answer was not exact.
The three special angles above allow us to express the trig ratios as simple fractions that represent the exact answer.
This means that when we have these special angles, we can work without a calculator. -
Activity 3: Example 1
Determine the value of the following WITHOUT a calculator:
- \( \sin 30^\circ \)
- \( \tan 45^\circ \)
- \( \cos 60^\circ \)
- \( \sin^2 60^\circ + \cos^2 60^\circ \)
Solution
When asked to determine the value of trig ratios without a calculator, you know you have to use the three special angles. It can help to make quick small sketches of the triangles with these angle in on the side of your page, like this.
- \( \sin 30^\circ = \displaystyle\frac{1}{2} \)
- \( \tan 45^\circ =1 \)
- \( \cos 60^\circ = \displaystyle\frac{1}{2}\)
- \( \sin^2 60^\circ + \cos^2 60^\circ =\displaystyle \left(\frac{\sqrt{3}}{2} \right)^2 + \displaystyle \left(\frac{1}{2} \right)^2 = \displaystyle\frac{3}{4}+\displaystyle\frac{1}{4}=1 \)
- \( \sin 30^\circ \)
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So far, we have had lots of practice calculating the values of different trig ratios. But why bother? What is the purpose of it all?
Trigonometry allows us to solve all sorts of problems in the real world. For example, a lighthouse is \( 80 \text{ m} \) tall. A ship out at sea measures the angle to the top of the lighthouse as \( 13.78^\circ \). How far away is the ship from the lighthouse?
Before we can solve this problem, there are a few things we need to know.
Firstly, an angle of elevation is an angle measured UP from the horizontal. In the situation above \( 13.78^\circ \) is the angle of elevation.
Secondly, an angle of depression is the angle measured DOWN from the horizontal.
Finally, we also need to know how to use our trig ratios to find the lengths of unknown sides.
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Activity 1: Finding the lengths of unknown sides
It’s time to start using trigonometry to solve real problems. One of the most commons types of problems you will encounter is to find the length of an unknown side in a right-angled triangle.
Have a look at the triangle below. Because it is a right-angled triangle and because we know the length of two sides, we can easily find the length of the third side by using Pythagoras’ Theorem.
Now look at the next triangle below. In this case, we only know the length of one of the sides so we cannot use Pythagoras. But we do have another piece of information. We know the size of one of the non-right angles.
With respect to the known angle, which trig ratio combines the length of the side we know (\( DE \)) and the length of \( x \), the side we are trying to find (\( EF \))?
With respect to \( \hat{\text{D}} \), we know the length of the adjacent side, and we want to find the length of the opposite side. Therefore, we can use tangent and say that \( \tan 34.16^\circ =\displaystyle \frac{\text{opp}}{\text{adj}}=\displaystyle \frac{EF}{DE}=\displaystyle \frac{x}{5.55} \). In other words, \( \tan 34.16^\circ =\displaystyle \frac{x}{5.55} \).
Now we have an equation that we can solve.
\( \begin{align*}\tan 34.16^\circ & =\displaystyle \frac{x}{5.55}\\\therefore x & =\tan 34.16^\circ \times 5.55\end{align*} \)
But what is the value of \( \tan 34.16^\circ \)? We can use any scientific calculator to find the value of any of the trig ratios for any angle. See the sequence of key presses required for different calculators. You will need to check which sequence your calculator uses.
Casio calculators (direct input): \( \boxed{\tan}\to \boxed{34.16}\to\boxed{)}\to\boxed{=} \)
Other calculators (indirect input):\( \boxed{34.16}\to\boxed{\tan}\to\boxed{=} \)
In either case, you should get that \( \tan 34.16^\circ =0.678579...\approx 0.679 \). Note that in this unit we will mostly show the Casio calculator key sequence.
Note In most situations the answers you get from your calculator for the value of a trig ratio are non-terminating, non-repeating decimals. For the greatest accuracy, round off only your final answerand round off to three decimal places unless told otherwise. Now we can solve for $latex \scriptsize x$ in our original equation.
\( \begin{align*}\tan 34.16{}^\circ & =\displaystyle \frac{x}{{5.55}}\\\therefore x & =\tan 34.16{}^\circ \times 5.55\\&=3.766\end{align*} \)
\( \boxed{\tan}\to \boxed{34.16}\to\boxed{)}\to\boxed{\times}\to\boxed{5.55}\to\boxed{=}\)
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Activity 2: Example 1
Use your calculator to calculate the following correct to three decimal places:
- \( \cos 58^\circ \)
- \( 2\times \sin 13.56^\circ \)
- \( \displaystyle \frac{{\cos 67.3^\circ }}{{\sin 41.97^\circ }}\)
Solution
- \( \boxed{\cos}\to\boxed{58}\to\boxed{)}\to\boxed{=}\to 0.530 \)
- \( \boxed{2}\to\boxed{\times}\to\boxed{\sin}\to\boxed{13.56}\to\boxed{)}\to\boxed{=}\to0.469 \)
- \( \boxed{\cos}\to\boxed{67.3}\to\boxed{)}\to\boxed{\div}\to\boxed{\sin}\to\boxed{41.97}\to\boxed{)}\to\boxed{=}\to0.577 \)
Note When calculating the value of trig ratios, make sure that your calculator is in DEGREE mode. You should see a small D or DEG symbol on screen. If you do not, consult your calculator’s manual for how to change back to DEGREE mode. -
Activity 3: Example 2
Find the length of (correct to three decimal places).
Solution
With respect to the angle that we have, we know the hypotenuse and we want the adjacent side. Cosine is the ratio that combines adjacent and hypotenuse.
\( \begin{align*}\cos 23.29{}^\circ & =\displaystyle \frac{x}{{4.46}}\\\therefore x & =4.46\times \cos 23.29{}^\circ \\ & =4.097\end{align*} \)
\( \boxed{4.46}\to\boxed{\times}\to\boxed{\cos}\to\boxed{23.29}\to\boxed{)}\to\boxed{=}\to4.097 \)If you would like to see some additional simple worked examples watch these two videos:
- SOHCAHTOA - Finding Missing Sides PART 1 (Duration: 04.37)
- SOHCAHTOA - Finding Missing Sides PART 2 (Duration: 02.34)
- SOHCAHTOA - Finding Missing Sides PART 1 (Duration: 04.37)
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Activity 4: Problems involving angles of elevation
Now we can return to our original lighthouse problem. Here it is again.
A lighthouse is \( 80 \text{ m} \)tall. A ship out at sea measures the angle to the top of the lighthouse a\( 13.78^\circ \). How far away is the ship from the lighthouse?
Solution
If you are not given a sketch of the situation, draw your own.
Relative to the angle that we have, we know the opposite side and we want the adjacent side. Tangent is the ratio that combines opposite and adjacent.
\( \begin{align*}\tan 13.78^\circ & =\displaystyle \frac{80}{x} \\\therefore x & =\displaystyle \frac{80}{\tan 13.78^\circ } \\ & =326.193\ \text{m}\end{align*}\)
Don’t forget to include the units (metres) in your final answer.
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Activity 5: Problems involving angles of depression
Here is a problem that involves an angle of depression.
An stuntman planning to ramp a motorcycle between two buildings stands at the top of one of the buildings that he knows is \( 35\ \text{m} \) height and measures an angle of depression to the foot of the other building of \( 53.24^\circ \). How far apart are the buildings?
Solution
Remember that if no sketch is given of the situation, make your own. Do this now on a piece of paper and then compare it to the one below.
We can then draw an even simpler diagram like this.
Notice how the angle of depression is \( 53.24^\circ \). Therefore, the angle inside the triangle is \( 36.76^\circ \) because these two angles add up to \( 90^\circ \). We can now solve for \( x \), the distance between the buildings.
Relative to the angle that we have, we know the adjacent side and we want the opposite side. Tangent is the ratio that combines opposite and adjacent.
\( \begin{align*}\tan 36.76^\circ & =\displaystyle \frac{x}{35} \\\therefore x & =35 \times \tan 36.76^\circ \\ & =26.15\ \text{m}\end{align*}\)
Don’t forget to include the units (metres) in your final answer.
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Activity 6: Additional online angle of elevation and depression examples and practice
- Angles of Elevation and Angle of Depression - Definition and Examples: Read more on angles of elevation and depression and get more practice working through example problems.
- Angles of elevation and depression: Work through additional angles of elevation and depression examples.
- Fly-by calibration: Play with a unique interactive simulation to get more practice using trig and angles of depression to solve real problems.
- Angles of Elevation and Angle of Depression - Definition and Examples: Read more on angles of elevation and depression and get more practice working through example problems.
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