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Precalculus for Team-Based Inquiry Learning PREVIEW Edition — Instructor Version

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Section 6.3 Trigonometric Ratios (TR3)

Subsection 6.3.1 Activities

Remark 6.3.1.

In this section, we will learn how to use right triangles to evaluate trigonometric functions. Before doing that, however, let’s review some key concepts of right triangles that can be helpful when solving.

Definition 6.3.2.

Given a right triangle with legs of length \(a\) and \(b\) and hypotenuse of length \(c\text{,}\) the following equation holds:
\begin{equation*} a^2+b^2=c^2 \end{equation*}
This is called the Pythagorean Theorem.

Remark 6.3.3.

The Pythagorean Theorem is helpful because if we know the lengths of any two sides of a right triangle, we can always find the length of the third side.

Activity 6.3.4.

Suppose two legs of a right triangle measure \(3\) inches and \(4\) inches.
(a)
Draw a picture of this right triangle and label the sides. Use \(x\) to refer to the missing side.
Answer.
Students should draw a triangle with \(x\) representing the hypotenuse and the two legs \(3\) and \(4\) inches respectively.
(b)
What is the value of \(x\) (i.e., the length of the third side)?
  1. \(5\) inches
  2. \(\sqrt{5}\) inches
  3. \(25\) inches
  4. \(16\) inches
  5. \(\sqrt{7}\) inches
Answer.
A

Activity 6.3.5.

Suppose the hypotenuse of a right triangle is \(13\) cm long and one of the legs is \(5\) cm long.
(a)
Draw a picture of this right triangle and label the sides. Use \(x\) to refer to the missing side.
Answer.
Figure 6.3.6. One example of how a student can draw the triangle.
(b)
What is the value of \(x\) (i.e., the length of the third side)?
  1. \(\displaystyle 144\)
  2. \(\displaystyle 12\)
  3. \(\displaystyle 194\)
  4. \(\displaystyle 14\)
Answer.
B

Definition 6.3.7.

Pythagorean triples are integers \(a\text{,}\) \(b\text{,}\) and \(c\) that satisfy the Pythagorean Theorem. Activity 6.3.4 and Activity 6.3.5 highlight some of the most common types of Pythagorean triples: \(3-4-5\) and \(5-12-13\text{.}\) All triangles similar to the \(3-4-5\) triangle will also have side lengths that are multiples of \(3-4-5\) (like \(6-8-10\)). Similarly, this is true for all triangles similar to the \(5-12-13\) triangle.

Activity 6.3.8.

Suppose you are given a right triangle where the hypotenuse is \(11\) cm long and one of the interior angles is \(60\)°.
(a)
Draw a picture of this right triangle and label the parts of the triangle given.
Answer.
Students should draw a right triangle with the hypotenuse labeled as \(11\) cm and one of the other angles (not the angle opposite the hypotenuse) is labeled \(60\)°.
(b)
What is the measure of the third angle?
  1. \(90\)°
  2. \(60\)°
  3. \(30\)°
  4. \(180\)°
Answer.
C
(c)
Suppose you are asked to find one of the sides of the right triangle. What additional information would you need to find the length of another side of the triangle?
Answer.
Students will probably notice that the Pythagorean Theorem is not helpful in this case because they only know the length of one side. This is a great opportunity to discuss how the Pythagorean Theorem is useful in finding side lengths when at least two sides are known.

Remark 6.3.9.

When working with right triangles, it is often helpful to refer to specific angles and sides. One way this is done is by using letters, such as \(A\) and \(a\) to show that these are an angle-side pair because every angle has a side opposite the angle in a triangle. Note that the capital letter indicates the angle, and the lower case letter indicates the side.
Another way to label the sides of a triangle is to use the relationships between a given angle within a triangle and the sides.
  • The hypotenuse of a right triangle is always the side opposite the right angle. This side also happens to be the longest side of the triangle.
  • The opposite side is the non-hypotenuse side across from a given angle.
  • The adjacent side is the non-hypotenuse side that is next to a given angle.
When given an angle, all sides of a triangle can be labeled from that angle’s perspective. For example, from angle \(A\)’s perspective, the sides of a right triangle are labeled as:
Figure 6.3.10. From the perspective of angle \(A\text{,}\) all sides of a right triangle can be labeled.

Definition 6.3.11.

We will define trigonometric functions for a given angle \(\theta\) as ratios between the side lengths of a right triangle. The first three trigonometric functions we will discuss are the sine function, the cosine function, and the tangent function.
\begin{gather*} \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\\ \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\\ \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \end{gather*}
Notice that these are defined according to the sides of a triangle from the perspective of an angle \(\theta\text{.}\) This is why it is important to be able to label the triangle correctly!

Activity 6.3.12.

Let \(ABC\) be a right triangle with lengths \(a=35\text{,}\) \(b=12\text{,}\) and \(c=37\text{.}\)
(a)
Draw a right triangle and label it with angles \(A\text{,}\) \(B\text{,}\) and \(C\text{,}\) with \(c\) being the hypotenuse of the triangle.
Answer.
Students should draw a triangle so that \(c\) is the hypotenuse and each angle is across from the correct measure.
(b)
Find \(\sin A\text{.}\)
  1. \(\displaystyle \dfrac{12}{37}\)
  2. \(\displaystyle \dfrac{35}{37}\)
  3. \(\displaystyle \dfrac{35}{12}\)
Answer.
B
(c)
Find \(\cos A\text{.}\)
  1. \(\displaystyle \dfrac{12}{37}\)
  2. \(\displaystyle \dfrac{35}{37}\)
  3. \(\displaystyle \dfrac{35}{12}\)
Answer.
A
(d)
Find \(\tan A\text{.}\)
  1. \(\displaystyle \dfrac{12}{37}\)
  2. \(\displaystyle \dfrac{35}{37}\)
  3. \(\displaystyle \dfrac{35}{12}\)
Answer.
C

Definition 6.3.13.

The three trigonometric ratios we have worked with so far (sine, cosine, and tangent) are referred to as the basic trigonometric functions. There are three additional functions, cosecant, secant, and cotangent that are found by taking the reciprocal of the basic trigonometric functions. These three ratios are referred to as the reciprocal trigonometric functions and can be defined as follows:
\begin{gather*} \csc\theta=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{1}{\sin\theta}\\ \sec\theta=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{1}{\cos\theta}\\ \cot\theta=\frac{\text{adjacent}}{\text{opposite}}=\frac{1}{\tan\theta} \end{gather*}

Activity 6.3.14.

Suppose you are given triangle \(ABC\text{,}\) where \(a=24\text{,}\) \(b=32\text{,}\) and \(c=40\text{,}\) with \(c\) being the hypotenuse of the triangle.
(a)
Find \(\sin{A}\text{.}\)
  1. \(\displaystyle \frac{4}{5}\)
  2. \(\displaystyle \frac{3}{5}\)
  3. \(\displaystyle \frac{5}{3}\)
  4. \(\displaystyle \frac{5}{4}\)
Answer.
B
(b)
Find \(\csc{A}\text{.}\)
  1. \(\displaystyle \frac{4}{5}\)
  2. \(\displaystyle \frac{3}{5}\)
  3. \(\displaystyle \frac{5}{3}\)
  4. \(\displaystyle \frac{5}{4}\)
Answer.
C
(c)
Find \(\cos{B}\text{.}\)
  1. \(\displaystyle \frac{4}{5}\)
  2. \(\displaystyle \frac{3}{5}\)
  3. \(\displaystyle \frac{5}{3}\)
  4. \(\displaystyle \frac{5}{4}\)
Answer.
A
(d)
Find \(\sec{B}\text{.}\)
  1. \(\displaystyle \frac{4}{5}\)
  2. \(\displaystyle \frac{3}{5}\)
  3. \(\displaystyle \frac{5}{3}\)
  4. \(\displaystyle \frac{5}{4}\)
Answer.
D
(e)
Find \(\cot{B}\text{.}\)
  1. \(\displaystyle \frac{4}{5}\)
  2. \(\displaystyle \frac{4}{3}\)
  3. \(\displaystyle \frac{5}{4}\)
  4. \(\displaystyle \frac{3}{4}\)
Answer.
D

Activity 6.3.15.

Suppose \(\cos\theta=\dfrac{12}{13}\text{.}\)
(a)
Draw a triangle and label one of the angles \(\theta\text{.}\) Then, label each side as "opposite," "adjacent," and "hypotenuse."
Answer.
Students should draw a triangle where the side across from \(\theta\) is labeled "opposite", the side next to the angle \(\theta\) is labeled "adjacent", and the "hypotenuse" as the third side across from the \(90\)-degree angle.
(b)
Use the given information to determine the length of each side of the triangle.
Answer.
Given that \(\cos\theta=\dfrac{12}{13}\text{,}\) we know that the side adjacent to \(\theta\) is \(12\) and the hypotenuse is \(13\text{.}\) Using the Pythagorean Theorem, the third side is \(5\text{.}\)
(c)
Find \(\csc\theta\text{.}\)
Answer.
\(\csc\theta=\dfrac{13}{5}\)
(d)
Find \(\cot\theta\)
Answer.
\(\cot\theta=\dfrac{12}{5}\)

Activity 6.3.16.

For each triangle given, determine which trigonometric function would be the most helpful in determining the length of the side of a triangle. Be sure to draw a picture of the triangle to help you determine the relationship between the given angle and sides. In each case assume angle \(C\) is the right angle.
(a)
In triangle \(ABC\text{,}\) \(B=37\)° and \(a=11\text{.}\) Which trigonometric function will best help determine the length of side \(c\text{?}\)
  1. sine
  2. cosine
  3. tangent
Answer.
B
(b)
In triangle \(ABC\text{,}\) \(A=32\)° and \(b=13\text{.}\) Which trigonometric function will best help determine the length of side \(a\text{?}\)
  1. sine
  2. cosine
  3. tangent
Answer.
C
(c)
In triangle \(ABC\text{,}\) angle \(A=24\)° and the hypotenuse has length \(14\text{.}\) Which trigonometric function will best help determine the length of side \(a\text{?}\)
  1. sine
  2. cosine
  3. tangent
Answer.
A

Activity 6.3.17.

In triangle \(ABC\text{,}\) \(B=37\)°, \(a=11\text{,}\) and \(c\) is the hypotenuse.
(a)
Draw a right triangle and label the angles/sides with the information given.
Answer.
Students should draw a right triangle where \(c\) is the hypotenuse, angle \(B\) is \(37\)°, and the side adjacent to \(b\) is \(11\text{.}\)
(b)
Which trigonometric equation will best help determine the length of side \(b\text{?}\)
  1. \(\displaystyle \sin{37^\circ} = \frac{b}{11}\)
  2. \(\displaystyle \tan{37^\circ} = \frac{b}{11}\)
  3. \(\displaystyle \cos{37^\circ} = \frac{b}{11}\)
Answer.
B
(c)
Use the equation from part (b) and solve for \(b\text{.}\) Round to the nearest tenth.
Hint.
Multiply by \(11\) on both sides and make sure you are in degree mode on your calculator.
Answer.
\(b = 8.3\)

Activity 6.3.18.

Sarah is standing \(500\) meters from the base of the Eiffel Tower. She looks at the top of the tower at an angle of \(31\)°.
(a)
Draw a diagram to represent the situation. Use \(x\) to refer to the missing side.
Answer.
Students should draw a right triangle where the \(31\)° is the angle formed with the ground, \(500\) meters as the side adjacent to the angle, and the height of the Eiffel Tower (side opposite \(31\)°) as \(x\text{.}\)
(b)
Which trig function could we use to find the height of the tower?
  1. sine
  2. cosine
  3. tangent
Answer.
C
(c)
How could we correctly set up the trigonometric function to find the height of the Eiffel Tower?
  1. \(\displaystyle \sin{31^\circ}=\frac{x}{500}\)
  2. \(\displaystyle \cos{31^\circ}=\frac{500}{x}\)
  3. \(\displaystyle \cos{31^\circ}=\frac{500}{x}\)
  4. \(\displaystyle \tan{31^\circ}=\frac{x}{500}\)
Answer.
D
(d)
Take your equation in part (c) and solve for \(x\text{.}\)
Answer.
\(x=500 \cdot \tan{31^\circ}\)
(e)
Find the height of the tower to the nearest hundredth.
  1. \(428.58\) meters
  2. \(300.43\) meters
  3. \(220.85\) meters
  4. \(257.52\) meters
Answer.
B

Exercises 6.3.2 Exercises

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