6. Trigonometric Ratios of Acute Angles
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Chapter 2
6.
Trigonometric Ratios of Acute Angles
This lesson delves into the relationship between trigonometric ratios and acute angles in right triangles. It explains how these ratios—sine, cosine, and tangent—can be used to find missing side lengths and angles. The text also introduces inverse trigonometric ratios as a tool for solving problems. For example, if you know the length of one side and the hypotenuse, you can find an unknown angle in the triangle. This knowledge is not just theoretical; it has practical applications in various fields like engineering, architecture, and even in everyday scenarios like calculating distances or heights.
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Lesson Settings & Tools
| | 17 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Trigonometric Ratios of Acute Angles
Slide of 17
In this lesson, the concept of similarity will be used to understand that side ratios in right triangles are properties of the angles. This will lead to the definition of trigonometric ratios for acute angles.
Catch-Up and Review
Here are a few recommended readings before getting started.
Try your knowledge on these topics.
a Select all the triangles that are similar to △ ABC.
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b Which of the following angles is a right angle?
{"type":"choice","form":{"alts":["∠ 1","∠ 2","∠ 3","∠ 4","∠ 5","∠ 6"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":3}
c Which of the following are right triangles?
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d Find the value of x by using the Pythagorean Theorem.
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Investigating a Side of a Right Triangle
The Leaning Tower of Pisa has a tilt of 4 degrees. Once, a worker maintaining it accidentally dropped a hammer from the top. The hammer landed 3 meters away from the base of the tower. Luckily, it did not hurt anyone!
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Comparing Ratios of Sides in Similar Right Triangles
Similar triangles have congruent angles and proportional sides. In the following applet, some of the ratios of the side lengths of two similar right triangles are compared. 
What conclusion can be made about the ratios of the side lengths of two similar triangles?
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Analyzing Ratios of Sides in Similar Right Triangles
Because all right angles are congruent, all right triangles have one pair of congruent angles. If they also have one pair of congruent acute angles, then the triangles have two pairs of congruent angles. Therefore, by the Angle-Angle Similarity Theorem, two triangles with one pair of congruent acute angles are similar.
Since corresponding sides of similar polygons are proportional, the ratios between corresponding sides of similar right triangles are the same.
AB/BC&=PQ/QR [1em] AC/BC&=PR/QR [1em] AC/AB&=PR/PQDiscussion
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Trigonometric Ratios
The ratios between side lengths of right triangles depend on the acute angles of the triangle. Some of these ratios receive a special name.
A trigonometric ratio relates two side lengths of a right triangle. Consider the right triangle △ ABC. One of its acute angles has been named θ.
Since it is opposite to the right angle, BC is the hypotenuse of the right triangle. The remaining sides — the legs — can be named relative to the marked angle θ. Because AB is next to ∠ θ, it is called the adjacent side. Conversely, because AC lies across from ∠ θ, it is called the opposite side.
The names of the three main ratios between side lengths are stated in the following table.
| Name | Definition | Notation |
|---|---|---|
| Sine of ∠ θ | Length of opposite side to ∠ θ/Hypotenuse | sin θ=opp/hyp |
| Cosine of ∠ θ | Length of adjacent side to ∠ θ/Hypotenuse | cos θ=adj/hyp |
| Tangent of ∠ θ | Length of opposite side to ∠ θ/Length of adjacent side to ∠ θ | tan θ=opp/adj |
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Understanding Trigonometric Ratios in Right Triangles
Dominika is helping Tadeo understand trigonometric ratios. She drew three right triangles for him to write trigonometric ratios with respect to the acute angle θ. Help Tadeo grasp this topic by selecting the correct answers!
a 
{"type":"choice","form":{"alts":["sin \u03b8=3\/5","sin \u03b8=4\/5","sin \u03b8=3\/4"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
b 
{"type":"choice","form":{"alts":["cos \u03b8=6\/10","cos \u03b8=8\/10","cos \u03b8=8\/6"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
c 
{"type":"choice","form":{"alts":["tan \u03b8=12\/5","tan \u03b8=12\/13","tan \u03b8=5\/13"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
Hint
a Identify the hypotenuse of the right triangle and the opposite side to ∠ θ.
b Identify the hypotenuse of the right triangle and the adjacent side to ∠ θ.
c Identify the opposite and adjacent sides to ∠ θ.
Solution
a The sine of ∠ θ is defined as the ratio of the length of the side opposite ∠ θ to the hypotenuse of the right triangle. Therefore, Tadeo needs to identify these two sides.
It can be seen above that the hypotenuse of the right triangle is 5 and that the length of the opposite side to ∠ θ is 3. With this information, the sine of ∠ θ can be written. sin θ = 3/5
b The cosine of ∠ θ is the ratio of the length of the adjacent side to ∠ θ to the hypotenuse of the right triangle. Therefore, Tadeo needs to identify these two sides.
The hypotenuse of the right triangle is 10 and the length of the adjacent side to ∠ θ is 6. With this information, the cosine of ∠ θ can be written. cos θ = 6/10
c The tangent of ∠ θ is the ratio of the length of the side opposite ∠ θ to the length of the side adjacent ∠ θ. Therefore, Tadeo needs to identify these two sides.
It can be seen above that the lengths of the opposite and adjacent sides to ∠ θ are 12 and 5, respectively. With this information, the tangent of ∠ θ can be written. tan θ = 12/5
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Explaining Trigonometric Ratios in Right Triangles
Despite the awesome explanations Dominika provided, Tadeo still does not get how to find trigonometric ratios. To help his friend, Dominika thought of one more exercise.
This time, Dominika drew one right triangle and stated its three side lengths. She also labeled one of the triangle's acute angles.
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Hint
Start by identifying the hypotenuse of the right triangle. Then identify the opposite and adjacent sides to ∠ θ. Finally, recall the definitions of sine, cosine, and tangent of an acute angle of a right triangle.
Solution
To find the sine, cosine, and tangent of ∠ θ, the hypotenuse of the right triangle needs to be identified. The opposite and adjacent sides to the angle also need to be identified.
It can be seen above that the hypotenuse is 17, and that the lengths of the opposite and adjacent sides to ∠ θ are 8 and 15, respectively. This information can be substituted into the definitions for sine, cosine, and tangent.
| Definition | Substitute |
|---|---|
| sin θ = Length of oppositeside to∠ θ/Hypotenuse | sin θ = 8/17 |
| cos θ = Length of adjacentside to∠ θ/Hypotenuse | cos θ = 15/17 |
| tan θ = Length of oppositeside to∠ θ/Length of adjacentside to∠ θ | tan θ = 8/15 |
Example
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Using Trigonometric Ratios to Find Side Lengths
Tadeo finally understands the topic! But wait, Dominika wants to level up and has let him know that trigonometric ratios can also be used to find missing side lengths of a right triangle. "Tell me more," Tadeo responds. An acute angle and the hypotenuse of a right triangle are given. To see whether Tadeo masters this topic, Dominika asked him to find the value of x, which is the length of the opposite side to the given angle.
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Hint
Identify the trigonometric ratio that should be used according to the given and desired lengths. Then, with the help of a calculator, set and solve an equation.
Solution
The hypotenuse of the right triangle is given, and the length of the opposite side to the given angle is to be found.
sin 60^(∘)=x/10
▼
Solve for x
MultEqn
LHS * 10=RHS* 10
sin 60^(∘) * 10=x/10 * 10
FracMultDenomToNumber
a/10* 10 = a
sin 60^(∘) * 10=x
Multiply
Multiply
10sin 60^(∘)=x
RearrangeEqn
Rearrange equation
x=10sin 60^(∘)
Degreein the third row.
Next the value of sin 60^(∘) can be calculated by pushing SIN followed by the angle measure.
x=10sin 60^(∘)
▼
UseCalc
Use a calculator
x=10(0.866025...)
Multiply
Multiply
x=8.660254...
RoundSigDig
Round to 3 significant digit(s)
x≈ 8.66
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Practice Finding Side Lengths Using Trigonometric Ratios
In the right triangles below, one acute angle and one side length are given. By using the corresponding trigonometric ratio, find the length of the side labeled x. Round the answer to one decimal place.
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Pythagorean Identities
By using trigonometric ratios, an important property of angles can be derived.
For any angle θ, the following trigonometric identities hold true.
sin^2 θ + cos^2 θ = 1
Proof
For Acute AnglesConsider a right triangle with a hypotenuse of 1.
By recalling the sine and cosine ratios, the lengths of the opposite and adjacent sides to ∠ θ can be expressed in terms of the angle.
| Definition | Substitute | Simplify | |
|---|---|---|---|
| sin θ | Length of oppositeside to∠ θ/Hypotenuse | opp/1 | opp |
| cos θ | Length of adjacentside to∠ θ/Hypotenuse | adj/1 | adj |
It can be seen that if the hypotenuse of a right triangle is 1, the sine of an acute angle is equal to the length of its opposite side. Similarly, the cosine of the angle is equal to the length of its adjacent side.
By the Pythagorean Theorem, the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Therefore, for the above triangle, the sum of the squares of sin θ and cos θ is equal to the square of 1.
sin^2 θ+ cos^2 θ&= 1^2 ⇓ & sin^2 θ+cos^2 θ&=1
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Using Trigonometry to Determine the Cosine of an Angle
The property seen before can be used, among other things, to find the sine or cosine ratio of an acute angle in a right triangle.
Kriz and his friends plan to spend Saturday afternoon playing video games. To optimize the space, they decide to tidy up the basement to ensure the console, snacks, and beverages are placed in the form of a right triangle. Kriz decides to set the snacks and the beverages 3 and 5 meters away from the console, respectively.
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Hint
The hypotenuse of the right triangle is 5 and the measure of the opposite side to ∠ θ is 3. With this information, the sine ratio can be found.
Solution
The hypotenuse of the right triangle is 5 and the measure of the opposite side to ∠ θ is 3. With this information, the sine ratio can be found.
sin θ =Length of opposite side toθ/Hypotenuse ⇓ sin θ = 3/5
This value can be substituted in the equation sin ^2 θ + cos ^2 θ =1.
Note that, when solving the equation for cos θ, only the principal root was considered. The reason is that the cosine of ∠ θ is the ratio between two side lengths, and side lengths are always positive. Therefore, the quotient is also positive.
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Using Inverse Trigonometric Ratios
Trigonometric ratios can also be used to find missing angles. Consider a right triangle △ ABC where the hypotenuse and a leg are given.
Suppose now that the measure of ∠ C is desired. Note that, apart from the hypotenuse, the side whose length is known is opposite to ∠ C. The trigonometric ratio that relates the hypotenuse and the opposite side to an acute angle in a right triangle is the sine ratio. sin ∠ C=length of oppositeside to ∠ C/hypotenuse ⇓ sin ∠ C=5/12 To find the measure of ∠ C, the inverse of the sine ratio could be used. sin ∠ C=5/12 ⇕ m ∠ C=sin ^(- 1)5/12
Finally, to find the value of sin ^(- 1) 512 and therefore the measure of ∠ C, a calculator will help. In the following example, it will be shown how to use a calculator to find the value of an inverse trigonometric ratio.Example
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Calculating Angles of Right Triangles
Previously, it was said that apart from being useful to find side lengths of a right triangle, trigonometric ratios can also be used to find missing angle measures.
Before playing video games with his friends, Kriz wants to finish his math homework to have a care-free weekend. He wants to find the measure of an acute angle in three different right triangles. By using the corresponding trigonometric ratios, help Kriz find m∠ θ in each triangle. Round the answer to the nearest degree.
a 
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b 
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c 
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Hint
a The lengths of the adjacent and opposite sides to ∠ θ are 12 and 35, respectively.
b The hypotenuse of the right triangle is 25 and the length of the adjacent side to ∠ θ is 7.
c The hypotenuse of the right triangle is 29 and the length of the opposite side to ∠ θ is 20.
Solution
a In the given diagram, it can be seen that the lengths of the opposite and adjacent sides to ∠ θ are 35 and 12, respectively. The trigonometric ratio that relates these two sides is the tangent ratio.
Thereofre, m∠ θ ≈ 71^(∘).
tan θ = length of oppositeside to∠ θ/length of adjacentside to∠ θ ⇓ tan θ = 35/12
To solve this equation, the inverse of the tangent function could be used. tan θ = 35/12 ⇕ m∠ θ =tan ^(- 1) 35/12
To find the value of tan ^(- 1) 3512, a calculator should to be used. First, the calculator must be set in degree mode. This is done by pushing MODE and selecting Degree
in the third row.
Next the value of tan ^(- 1) 3512, can be calculated by pushing 2ND, followed by TAN, and 35/12.
b In the given diagram, it is shown that the length of the adjacent side to ∠ θ is 7 and that the hypotenuse of the right triangle is 25. The trigonometric ratio that relates these two sides is the cosine ratio.
It was found that m∠ θ ≈ 74^(∘).
cos θ = length of adjacentside to∠ θ/hypotenuse ⇓ cos θ = 7/25
To solve this equation, the inverse of the cosine function is needed. cos θ = 7/25 ⇕ m∠ θ =cos ^(- 1) 7/25
To find the value of cos ^(- 1) 725, a calculator should be used. Just like before, the calculator must be set in degree mode. This is done by pushing MODE and selecting Degree
in the third row.
Next the value of cos ^(- 1) 725, is calculated by pushing 2ND, followed by COS, and 7/25.
c In the diagram, it can be seen that the hypotenuse of the right triangle is 29 and that the length of the opposite side to ∠ θ is 20. The trigonometric ratio that relates these two sides is the sine ratio.
Thereofre, m∠ θ ≈ 44^(∘).
sin θ = length of oppositeside to∠ θ/hypotenuse ⇓ sin θ = 20/29
To solve this equation, the inverse of the sine function can be used. sin θ = 20/29 ⇕ m∠ θ =sin ^(- 1) 20/29
To find the value of sin ^(- 1) 2029, a calculator should be used. Just like in Parts A and B, the calculator must be set in degree mode by pushing MODE and selecting Degree
in the third row.
Next the value of sin ^(- 1) 2029, can be calculated by pushing 2ND, followed by SIN, and 20/29.
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Practice Finding Angles Using Trigonometric Ratios
In the following right triangles, two side lengths are given. By using the corresponding trigonometric ratio, find m∠ θ. Round the answer to nearest degree.
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Reciprocal Trigonometric Ratios
Apart from the sine, cosine, and tangent ratios, there are three other trigonometric ratios that are worth mentioning.
Consider the right triangle △ ABC.
The so called reciprocal ratios are written in the next table.
| Name | Definition | Notation |
|---|---|---|
| Cosecant of ∠ θ | Hypotenuse/Length of opposite side to ∠ θ | csc θ=hyp/opp |
| Secant of ∠ θ | Hypotenuse/Length of adjacent side to ∠ θ | sec θ=hyp/adj |
| Cotangent of ∠ θ | Length of adjacent side to ∠ θ/Length of opposite side to ∠ θ | cot θ=adj/opp |
These ratios can be defined in terms of sine, cosine, and tangent.
Rule
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Reciprocal Identities
The trigonometric ratios cosecant, secant, and cotangent are reciprocals of sine, cosine, and tangent, respectively.
csc θ=1/sin θ
sec θ=1/cos θ
cot θ=1/tan θ
Proof
Consider a right triangle with the three sides labeled with respect to an acute angle θ.
sin θ=opp/hyp
▼
Solve for 1/sin θ
MultEqn
LHS * 1/sin θ=RHS* 1/sin θ
1=opp/hyp (1/sin θ)
MultEqn
LHS * hyp/opp=RHS* hyp/opp
hyp/opp=1/sin θ
RearrangeEqn
Rearrange equation
1/sin θ=hyp/opp
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Finding Reciprocal Identities
If the sine, cosine, and tangent ratios are known, then their reciprocals cosecant, secant, and cotangent can be calculated without too much effort.
LaShay is really good at her favorite subject, Geometry. She has been appointed by Jefferson High's principal to do some tutoring for some of her classmates after school. To do so, she drew a right triangle. She then asked her peers to find all six trigonometric ratios with respect to the marked angle θ.
Help LaShay's classmates find the trigonometric ratios!
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Hint
Identify the hypotenuse of the right triangle and the opposite and adjacent sides to ∠ θ.
Solution
The hypotenuse of the right triangle and the opposite and adjacent sides to ∠ θ will be identified.
It can be seen that the hypotenuse is 101 and the lengths of the opposite and adjacent sides to ∠ θ are 20 and 99, respectively. With this information, the sine, cosine, and tangent ratios can be found. sin θ &= 20/101 [0.8em] cos θ &= 99/101 [0.8em] tan θ &= 20/99 The reciprocals of the above ratios are the cosecant, secant, and cotangent of ∠ θ. sin θ = 20/101 & ⇒ csc θ = 101/20 [0.9em] cos θ = 99/101 & ⇒ sec θ = 101/99 [0.9em] tan θ = 20/99 & ⇒ cot θ = 99/20
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Calculating a Side of a Right Triangle
With the topics learned in this lesson, the challenge presented at the onset can now be solved. Previously, it was learned that the Leaning Tower of Pisa has a tilt of 4 degrees. The hammer dropped by the worker landed 3 meters away from the base of the tower.
With the given information, the vertical distance traveled by the hammer can be calculated. Write the answer to the nearest tenth. 
The trigonometric ratio that relates an angle of a right triangle with its opposite and adjacent sides is the tangent ratio.
tan θ = length of opposite side to ∠ θ/length of adjacent side to ∠ θ ⇓ tan 86^(∘)=x/3
This equation can be solved for x, which is the vertical distance traveled by the hammer.
The distance traveled by the hammer is about 42.9 meters.
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Hint
Draw a right triangle and identify the given information.
Solution
If the tower has a tilt of 4^(∘), then the acute angle formed by the ground and the tower itself is the difference between 90^(∘) and 4^(∘). 90^(∘)- 4^(∘)= 86^(∘) A right triangle can be drawn with an acute angle whose measure is 86^(∘). Furthermore, the length of the adjacent side to this angle is 3 meters, and the length of its opposite side is unknown.
tan 86^(∘)=x/3
▼
Solve for x
MultEqn
LHS * 3=RHS* 3
tan 86^(∘)* 3=x
RearrangeEqn
Rearrange equation
x=tan 86^(∘)* 3
UseCalc
Use a calculator
x=42.901998...
RoundDec
Round to 1 decimal place(s)
x≈ 42.9
Trigonometric Ratios of Acute Angles
Exercise 3.1
Below we see two circles inscribed in an equilateral triangle.
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From the exercise, we know that the triangle is equilateral. In such a triangle each angle has a measure of 60^(∘).
Since the triangle is equilateral, an altitude will always bisect the angle it is drawn from. Let's draw the altitude from the top angle.
Notice that both circles are inscribed in the triangle. According to the Tangent to a Circle Theorem, the circles and the triangle intersect at right angles at their points of tangency. With this information, we can draw the radius of the larger circle thereby forming a 30-60-90 triangle where we know the length of the shorter leg and an expression for the hypotenuse.
Let’s focus solely on the triangle for now.
Since we know an angle, the hypotenuse, and the angle's opposite leg, we can use the sine ratio to solve for x.
As we can see, the diameter of the smaller circle is 1.32 units.
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The manufacturer will test the elasticity of every 20^(th) rubber band they produce. They will execute the test by attaching the ends of each rubber band between two walls, then hanging a weight in the middle of the band, causing it to stretch.
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Let's label the length of the unstretched rubber band x. When the weight is applied, the rubber band is stretched making an isosceles triangle. The legs of this triangle are each half of x multiplied by a certain stretch factor, which we label a. Therefore, each leg has a length of a(0.5x).
If we draw the altitude from the triangle's vertex angle, we get two right triangles where the expressions for a leg and the hypotenuse are known.
The measure of ∠ v can, at most, be 35^(∘). If we substitute this angle and the expressions for the leg and the hypotenuse into the cosine ratio, we can determine a.
As we can see, the stretch factor a is approximately 1.22 when the angle is 35^(∘). This factor corresponds to a length increase of 22 %. This is the percentage the band can stretch before being thrown away.
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Trigonometric Ratios of Acute Angles
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