Right Triangle Trigonometry

Section 1.4 Right Triangle Trigonometry

During a voyage, a navigator utilizes a reference course —a line connecting the starting point and destination—to monitor their position. When the waʻa (canoe) encounters winds that veer it off course, the navigator mentally plots their position relative to the reference course. To ensure the destination isn’t missed, navigators must monitor their deviation from the intended course, involving measurement of the angle of deviation from the reference course (in units of houses) and determining the distance traveled. This section explores the calculation of trigonometric functions using right triangles, enabling us to assess how much the waʻa has strayed from its intended reference course.

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Subsection 1.4.1 Trigonometric Ratios

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Definition 1.4.1. Trigonometric Ratios.

Consider a right triangle with

θ

as one of its acute angles. The trigonometric ratios are defined as follows:

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\begin{aligned} \sin θ & = \frac{opposite}{hypotenuse} & \cos θ & = \frac{adjacent}{hypotenuse} & \tan θ & = \frac{opposite}{adjacent} \\ \csc θ & = \frac{hypotenuse}{opposite} & \sec θ & = \frac{hypotenuse}{adjacent} & \cot θ & = \frac{adjacent}{opposite} \end{aligned}

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A common mnemonic for remembering these relationships is SOHCAHTOA, formed from the first letters of “Sine is Opposite over Hypotenuse, Cosine is Adjacent over Hypotenuse, Tangent is Opposite over Adjacent.”

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Based on the definition of the six trigonometric functions, we have the following trigonometric identities.

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Definition 1.4.2. Reciprocal Identities.

\begin{aligned} \sin θ & = \frac{1}{\csc θ} & \cos θ & = \frac{1}{\sec θ} & \tan θ & = \frac{1}{\cot θ} \\ \csc θ & = \frac{1}{\sin θ} & \sec θ & = \frac{1}{\cos θ} & \cot θ & = \frac{1}{\tan θ} \end{aligned}

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Definition 1.4.3. Quotient Identities.

\begin{aligned} \tan θ & = \frac{\sin θ}{\cos θ} & \cot θ & = \frac{\cos θ}{\sin θ} \end{aligned}

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Example 1.4.4.

Find the exact values of the six trigonometric ratios of the angle

θ

in the given triangle.

Solution.

By the definition of the trigonometric ratios, we have

\begin{aligned} \sin θ & = \frac{3}{5} & \cos θ & = \frac{4}{5} & \tan θ & = \frac{3}{4} \\ \csc θ & = \frac{5}{3} & \sec θ & = \frac{5}{4} & \cot θ & = \frac{4}{3} \end{aligned}

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Example 1.4.5.

Find the exact values of the six trigonometric ratios of the angle

θ

in the given triangle.

Solution.

Notice that

θ

is in a different position. Here, the adjacent side is 3 and the opposite side is 5. If we let

h

denote the hypotenuse, then we can use the Pythagorean Theorem to get

h = \sqrt{5^{2} + 2^{2}} = \sqrt{29}

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Then by the definition of the trigonometric ratios, we have

\begin{aligned} \sin θ & = \frac{5}{\sqrt{29}} = \frac{5 \sqrt{29}}{29} & \cos θ & = \frac{2}{\sqrt{29}} = \frac{2 \sqrt{29}}{29} & \tan θ & = \frac{5}{2} \\ \csc θ & = \frac{\sqrt{29}}{5} & \sec θ & = \frac{\sqrt{29}}{2} & \cot θ & = \frac{2}{5} \end{aligned}

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Subsection 1.4.2 Special Triangles

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The angles

30^{\circ}, 45^{\circ}, 60^{\circ}

(

\frac{π}{6}, \frac{π}{4}, \frac{π}{3}

) give special values for trigonometric functions. The following figures are used to calculate trigonometric values.

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The trigonometric values for the special angles

0, 30^{\circ}, 45^{\circ}, 60^{\circ}, 90^{\circ}

(0, \frac{π}{6}, \frac{π}{4}, \frac{π}{3}, \frac{π}{2})

are given in Table 1.4.6.

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Table 1.4.6. Values of the trigonometric functions in Quadrant I


$θ$	$θ$	$\sin θ$	$\cos θ$	$\tan θ$
(degrees)	(radians)

$0^{\circ}$	0	0	1	0

$30^{\circ}$	$\frac{π}{6}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$

$45^{\circ}$	$\frac{π}{4}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	1

$60^{\circ}$	$\frac{π}{3}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$

$90^{\circ}$	$\frac{π}{2}$	1	0	undefined

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Subsection 1.4.3 Cofunctions

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The symmetry between

\sin θ

and

\cos θ

becomes evident when reversing the order of sine and cosine values from

90^{\circ}

0^{\circ} .

This symmetry yields

\sin 0^{\circ} = \cos 90^{\circ},

\sin 30^{\circ} = \cos 60^{\circ},

\sin 45^{\circ} = \cos 45^{\circ},

\sin 60^{\circ} = \cos 30^{\circ},

and

\sin 90^{\circ} = \cos 0^{\circ} .

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This pattern between sine and cosine is no coincidence; it emerges because the three angles in a triangle add up to

180^{\circ}

π

radians. When considering a right triangle, the remaining two angles combine to form

90^{\circ}

\frac{π}{2}

radians, making them complementary angles.

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Consider the right triangle in the figure above, where angles

α

and

β

are complementary angles. Side

a

is opposite of angle

α,

and side

b

is opposite of angle

β .

Notice that we can also describe side

b

as adjacent to angle

α

and side

a

as adjacent to angle

β .

Therefore,

\sin α = \frac{opposite}{hypotenuse} = \frac{a}{c} ~and~ \cos β = \frac{adjacent}{hypotenuse} = \frac{a}{c}

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Thus we can conclude that

\sin α = \frac{a}{c} = \cos β

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Sine and cosine are called cofunctions because of this relationship between these functions and their complementary angles. We can obtain similar relationships for all trigonometric functions:

\begin{aligned} \sin α = \frac{a}{c} & = \cos β & \cos α = \frac{b}{c} & = \sin β & \tan α = \frac{a}{b} & = \cot β \\ \csc α = \frac{c}{a} & = \sec β & \sec α = \frac{c}{b} & = \csc β & \cot α = \frac{b}{a} & = \tan β \end{aligned}

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Since

α

and

β

are complementary angles,

α + β = 90^{\circ} .

Rearranging, we get

β = 90^{\circ} - α .

Substituting this into our cofunctions and replacing

α

with

θ,

we get our cofunction identities.

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Definition 1.4.7. Cofunction Identities.

The cofunction identities in degrees are

\begin{aligned} \sin θ & = \cos (90^{\circ} - θ) & \cos θ & = \sin (90^{\circ} - θ) & \tan θ & = \cot (90^{\circ} - θ) \\ \csc θ & = \sec (90^{\circ} - θ) & \sec θ & = \csc (90^{\circ} - θ) & \cot θ & = \tan (90^{\circ} - θ) \end{aligned}

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The cofunction identities in radians are

\begin{aligned} \sin θ & = \cos (\frac{π}{2} - θ) & \cos θ & = \sin (\frac{π}{2} - θ) & \tan θ & = \cot (\frac{π}{2} - θ) \\ \csc θ & = \sec (\frac{π}{2} - θ) & \sec θ & = \csc (\frac{π}{2} - θ) & \cot θ & = \tan (\frac{π}{2} - θ) \end{aligned}

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Example 1.4.8.

The Cofunction Identities explains the symmetry in Table 1.4.6 .

\begin{aligned} \sin 0^{\circ} & = \cos (90^{\circ} - 0^{\circ}) = \cos 90^{\circ} & \sin 60^{\circ} & = \cos (90^{\circ} - 60^{\circ}) = \cos 30^{\circ} \\ \sin 30^{\circ} & = \cos (90^{\circ} - 30^{\circ}) = \cos 60^{\circ} & \sin 90^{\circ} & = \cos (90^{\circ} - 90^{\circ}) = \cos 0^{\circ} \\ \sin 45^{\circ} & = \cos (90^{\circ} - 45^{\circ}) = \cos 45^{\circ} \end{aligned}

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Remark 1.4.9. Patterns in the Trigonometric Table.

Learning the values of the trigonometric functions in this table can increase your confidence and efficiency in trigonometry. To help remember the values of sine and cosine, we utilize cofunctions and also write them in the form

\sqrt{\cdot} / 2


$θ$	$θ$	$\sin θ$	$\cos θ$

$0$	$0^{\circ}$	$\sqrt{0} / 2$	$\sqrt{4} / 2$

$π / 6$	$30^{\circ}$	$\sqrt{1} / 2$	$\sqrt{3} / 2$

$π / 4$	$45^{\circ}$	$\sqrt{2} / 2$	$\sqrt{2} / 2$

$π / 3$	$60^{\circ}$	$\sqrt{3} / 2$	$\sqrt{1} / 2$

$π / 2$	$90^{\circ}$	$\sqrt{4} / 2$	$\sqrt{0} / 2$

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which simplifies to the values in Table 1.4.6.

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Subsection 1.4.4 Using a Calculator

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Sometimes you may encounter an angle other than the special angles described above. In this case, you will have to use a calculator.

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First, be sure that your angle is either in degrees or radians, depending on the problem, refer to your calculator’s manual for instructions. Most calculators will have a special button for the sine, cosine, and tangent functions. Depending on your calculator, you may see the following keys for

Function	Calculator Key

sine	SIN
cosine	COS
tangent	TAN

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To calculate cosecant, secant, and cotangent, you will need to use the identity

\begin{aligned} \csc θ & = \frac{1}{\sin θ}, & \sec θ & = \frac{1}{\cos θ}, & \cot θ & = \frac{1}{\tan θ} \end{aligned}

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Answers produced by calculators are estimates and you should pay close attention to see if the question is asking for the exact solution or a decimal approximation. For example, if you need to calculate

\sin 45^{\circ} = \frac{\sqrt{2}}{2},

the calculator may give the answer as

\sin 45^{\circ} \approx 0.70710678,

which is a decimal approximation since the actual solution goes on forever. Unless stated otherwise, answers in the book should be exact, e.g.

\frac{\sqrt{2}}{2}

and not 0.70710678.

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Example 1.4.10.

Use a calculator to evaluate

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(a)

\sin 22^{\circ}

Solution.

Before proceeding, we confirm that our calculator is set to either degree or radian mode. Additionally, for the sake of simplicity, we will round our answers to four decimal places.

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Input: SIN (22); Output:

0.3746

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(b)

\cos 5^{\circ}

Solution.

Input: COS (5); Output:

0.9962

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(c)

\cot 53^{\circ}

Solution.

Since most calculators do not have a key for cotangent, we Input: (1/ TAN (53)); Output:

\frac{1}{1.3270} \approx 0.7536

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(d)

\cos 5 rad

Solution.

Since this problem uses radians, we must change the mode on our calculator then Input: COS (5); Output:

0.2837 .

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Remark 1.4.11.

Observe that

\cos 5^{\circ} \neq \cos 5 rad .

This emphasizes the significance of verifying whether your calculator is in degree or radian mode.

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Subsection 1.4.5 Solving Right Triangles

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Consider the following right triangle where side

a

is opposite angle

α,

side

b

is opposite angle

β,

and side

c

is the hypotenuse. Since

α

and

β

are complementary angles, we have

α + β = 90^{\circ}

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Additionally, by the Pythagorean Theorem, we have

a^{2} + b^{2} = c^{2}

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Definition 1.4.12.

To solve a triangle is the process of determining the values for all three lengths of its sides and the measures of all three angles, based on provided information about the triangle.

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Remark 1.4.13. Solving Right Triangles.

In solving a right triangle, the following relationships are useful:

α + β = 90^{\circ}, a^{2} + b^{2} = c^{2}

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Example 1.4.14.

Solve the right triangle. Round your answer to two decimal places.

Solution.

Given that this is a right triangle, we already know one angle is

90^{\circ},

and we have an additional angle of

50^{\circ}

along with an adjacent side length of 16. To solve this triangle, we need to determine the values of sides

a,

c,

and

β .

We begin by finding the measure of angle

β .

Since

50^{\circ} + β = 90^{\circ}

we have

β = 90^{\circ} - 50^{\circ} = 40^{\circ}

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Next, we will solve for side

a .

Using the angle

50^{\circ},

where the adjacent side is 16 and side

a

is the side opposite to the angle, we can apply the tangent function, which relates the opposite and adjacent sides:

\tan 50^{\circ} = \frac{a}{16}

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Multiplying both sides by 16 we get

a = 16 \cdot \tan 50^{\circ} \approx 19.07

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Using the Pythagorean Theorem, we get

c^{2} = 16^{2} + {19.07}^{2} \approx 619.66

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Thus

c \approx \sqrt{619.66} \approx 24.89

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Subsection 1.4.6 Solving Applied Problems

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Example 1.4.15. Deviation.

We are now ready to calculate the deviation example proposed at the start of this section. In an average day of sailing, a waʻa sails for 120 nautical miles (NM). If Hikianalia is supposed to sail in the direction of Hikina (East), but currents have deviated her course by one house so she actually sailed in the house Lā, how far off the course has Hikianalia deviated?

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Solution.

From the Star Compass (Figure 1.1.4), the house Lā is one house (

{11.25}^{\circ}

) from Hikina. If we let

y

denote the distance deviated from the reference course, our right triangle becomes:

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Since we know the hypotenuse of the triangle and want to find the side opposite of the angle, we will use the sine function:

\sin {11.25}^{\circ} = \frac{opposite}{hypotenuse} = \frac{y}{120 NM}

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multiplying both sides by 120, we get

y = 120 \cdot \sin {11.25}^{\circ} NM \approx 23.4 NM

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So Hikianalia has deviated 23.4 nautical miles north from the reference course.

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Example 1.4.16. Solar panels.

Solar panels harness the sun’s energy to generate electricity, and for optimal energy output, they should be oriented perpendicularly to the sun’s light. The sun’s angle of elevation varies based on latitude, and in Hawai‘i, for instance, south-facing solar panels are recommended to have a pitch of

21^{\circ}

to align with the sun’s rays. When installing a solar panel, determining its pitch might pose challenges. Instead of measuring the angle directly, an alternative approach involves measuring the height of the panel’s top. What height should a south-facing solar panel, measuring 65 inches in length, be installed at to achieve the desired angle of

21^{\circ} ?

Round your answer to the nearest tenth of an inch.

Solution.

We begin by drawing the triangle.

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Since we know the desired angle of pitch of the solar panel and the length of the panel, we can set up the following equation

\begin{aligned} \sin 21^{\circ} & = & \frac{opposite}{hypotenuse} = \frac{height}{65 in} \\ height & = & 65 in \cdot \sin 21^{\circ} \approx 23.3 in \end{aligned}

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Thus, when installing a solar panel in Hawai‘i, the top of the solar panel should be positioned 23.3 inches above the bottom to optimize energy output.

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Exercises 1.4.7 Exercises

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Exercise Group.

Find the exact values of the six trigonometric functions of the angle

θ

in each triangle.

Section 1.4 Right Triangle Trigonometry

Subsection 1.4.1 Trigonometric Ratios

Definition 1.4.1. Trigonometric Ratios.

Definition 1.4.2. Reciprocal Identities.

Definition 1.4.3. Quotient Identities.

Example 1.4.4.

Solution.

Example 1.4.5.

Solution.

Subsection 1.4.2 Special Triangles

Subsection 1.4.3 Cofunctions

Definition 1.4.7. Cofunction Identities.

Example 1.4.8.

Remark 1.4.9. Patterns in the Trigonometric Table.

Subsection 1.4.4 Using a Calculator

Example 1.4.10.

(a)

Solution.

(b)

Solution.

(c)

Solution.

(d)

Solution.

Remark 1.4.11.

Subsection 1.4.5 Solving Right Triangles

Definition 1.4.12.

Remark 1.4.13. Solving Right Triangles.

Example 1.4.14.

Solution.

Subsection 1.4.6 Solving Applied Problems

Example 1.4.15. Deviation.

Solution.

Example 1.4.16. Solar panels.

Solution.

Exercises 1.4.7 Exercises

Exercise Group.

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Exercise Group.

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Exercise Group.

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Exercise Group.

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