1.2: Right Triangle Trigonometry (2024)

Learning Objectives

• Define the six trigonometric functions in terms of \(x,y,r\) (rectangular coordinates) and language associated with right triangles.
• Find function values for 30°(\(\dfrac{\pi}{6}\)),45°(\(\dfrac{\pi}{4}\)),and 60°(\(\dfrac{\pi}{3}\)).
• Use equal cofunctions of complementary angles.
• Use the definitions of trigonometric functions of any angle.
• Use right-triangle trigonometry to solve applied problems.

The Six Trigonometric Functions

For the point (\(x\), \(y\)) on a circle of radius \(r\) at an angle of \(\theta\), we can define the six trigonometricfunctions as the ratios of the sides of the corresponding triangle:

The sine function: \(\sin (\theta )=\dfrac{y}{r}\)

The cosine function: \(\cos (\theta )=\dfrac{x}{r}\)

The tangentfunction: \(\tan(\theta )=\dfrac{y}{x}\)

The cosecantfunction: \(\csc(\theta )=\dfrac{r}{y}\)

The secantfunction: \(\sec(\theta )=\dfrac{r}{x}\)

The cotangentfunction: \(\cot(\theta )=\dfrac{x}{y}\)

Example \(\PageIndex{1}\)

The point (3, 4) is on the circle of radius 5 at some angle \(\theta\). Find the six trigonometric function values of \(\theta\)

Solution

We have \(x=3\), \(y=4\), and \(r=5\). Using the previously listed definitions we have

\[\sin (\theta )=\dfrac{y}{r} =\dfrac{4}{5}\\ \cos (\theta )=\dfrac{x}{r} =\dfrac{3}{5}\\ \tan(\theta )=\dfrac{y}{x} =\dfrac{4}{3}\\ \csc(\theta )=\dfrac{r}{y} =\dfrac{5}{4}\\ \sec(\theta )=\dfrac{r}{x} =\dfrac{5}{3}\\ \cot(\theta )=\dfrac{x}{y} =\dfrac{3}{4} \]

Using Right Triangles to Evaluate Trigonometric Functions

We now define the six trigonometric functions in terms of the names of the sides of a right triangle. First, let'srecreate our right triangle (shown inFigure \(\PageIndex{2}\).If we drop a vertical line segment from the point \((x,y)\) to the x-axis, we have a right triangle whose vertical side has length \(y\) and whose horizontal side has length \(x\). We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.

We want to have another way to express the ratios without relying on a set of axes. To be able to use these ratios freely, we will give the sides more general names: Instead of \(x\),we will call the side between the given angle and the right angle the adjacent side to angle \(\theta\). (Adjacent means “next to.”) Instead of \(y\),we will call the side most distant from the given angle the opposite side from angle \(\theta\). And instead of \(r\),we will call the side of a right triangle opposite the right angle the hypotenuse. These sides are labeled in Figure \(\PageIndex{3}\).

Understanding Right Triangle Relationships

Given a right triangle with an acute angle of \(\theta\),

\[\begin{align} \sin (\theta) &= \dfrac{\text{opposite}}{\text{hypotenuse}} \label{sindef}\\ \cos (\theta) &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \label{cosdef}\\ \tan (\theta) &= \dfrac{\text{opposite}}{\text{adjacent}} \label{tandef} \\ \csc(\theta) &= \dfrac{\text{hypotenuse}}{\text{opposite}} \label{cscdef}\\\sec(\theta) &= \dfrac{\text{hypotenuse}}{\text{adjacent}} \label{secdef}\\ \cot(\theta) &= \dfrac{\text{adjacent}}{\text{opposite}} \label{cotdef}\\\end{align}\]

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.” To get the other three, we remember that cosecant is the reciprocal of sine, secant is the reciprocal of cosine, and cotangent is the reciprocal of tangent.

Example \(\PageIndex{2}\): Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure \(\PageIndex{4}\), find the six trigonometric function values of angle \(\alpha\).

Solution

From the perspective of angle \(\alpha\), the adjacent side has length 15, the opposite side has length 8, and the hypotenuse has length 17, so we have

\[\begin{align*} \sin(\alpha) &= \dfrac{\text{O}}{\text{H}} = \dfrac{8}{17} \\ \cos (\alpha) &= \dfrac{\text{A}}{\text{H}}= \dfrac{15}{17} \\\tan(\alpha) &= \dfrac{\text{O}}{\text{A}} = \dfrac{8}{15} \\\csc(\alpha) &= \dfrac{\text{H}}{\text{O}}= \dfrac{17}{8} \\\sec(\alpha) &= \dfrac{\text{H}}{\text{A}}= \dfrac{17}{15} \\\cot(\alpha) &= \dfrac{\text{A}}{\text{O}} = \dfrac{15}{8} \end{align*}\]

Exercise \(\PageIndex{1}\)

Given the triangle shown in Figure \(\PageIndex{5}\), find the value of \(\sin t\).

Answer

\(\frac{7}{25}\)

Relating Angles and Their Functions

When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure \(\PageIndex{6}\). The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.

Example \(\PageIndex{3}\): Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure \(\PageIndex{7}\), evaluate \( \sin α, \cos α, \tan α, \sec α, \csc α,\) and \( \cot α\).

Solution

\[ \begin{align*} \sin α &= \dfrac{\text{opposite } α}{\text{hypotenuse}} = \dfrac{4}{5} \\ \cos α &= \dfrac{\text{adjacent to }α}{\text{hypotenuse}}=\dfrac{3}{5} \\ \tan α &= \dfrac{\text{opposite }α}{\text{adjacent to }α}=\dfrac{4}{3} \\ \sec α &= \dfrac{\text{hypotenuse}}{\text{adjacent to }α}= \dfrac{5}{3} \\ \csc α &= \dfrac{\text{hypotenuse}}{\text{opposite }α}=\dfrac{5}{4} \\ \cot α &= \dfrac{\text{adjacent to }α}{\text{opposite }α}=\dfrac{3}{4} \end{align*}\]

Exercise \(\PageIndex{2}\)

Using the triangle shown in Figure \(\PageIndex{8}\), evaluate \( \sin t, \cos t,\tan t, \sec t, \csc t,\) and \(\cot t\).

Answer

\[\begin{align*} \sin t &= \frac{33}{65}, \cos t= \frac{56}{65},\tan t= \frac{33}{56}, \\ \\ \sec t &= \frac{65}{56},\csc t= \frac{65}{33},\cot t= \frac{56}{33} \end{align*}\]

Finding Trigonometric Functions of Special Angles Using Side Lengths

We nowuse our right triangle definitions to evaluate trigonometric values ofspecial angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. Our special angles are\(30°, 45°,\) and \(60°\)

Recall from geometry the \(30°,60°,90°\) triangle, which can also be described as a \(\frac{π}{6}, \frac{π}{3},\frac{π}{2}\) triangle. The sides have lengths in the relation \(s,\sqrt{3}s,2s.\) The sides of a \(45°,45°,90° \)triangle, which can also be described as a \(\frac{π}{4},\frac{π}{4},\frac{π}{2}\) triangle, have lengths in the relation \(s,s,\sqrt{2}s.\) These relations are shown in Figure \(\PageIndex{9}\).

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

Using Equal Cofunction of Complements

If we look more closely at the relationship between the sine and cosine of the special angles relative to the unit circle, we will notice a pattern. In a right triangle with angles of \(\frac{π}{6}\) and \(\frac{π}{3}\), we see that the sine of \(\frac{π}{3}\), namely \(\frac{\sqrt{3}}{2}\), is also the cosine of \(\frac{π}{6}\), while the sine of \(\frac{π}{6}\), namely \(\frac{1}{2},\) is also the cosine of \(\frac{π}{3}\) (Figure \(\PageIndex{10}\)).

\[\begin{align*} \sin \frac{π}{3} &= \cos \frac{π}{6}=\frac{\sqrt{3}s}{2s}=\frac{\sqrt{3}}{2} \\ \sin \frac{π}{6} &= \cos \frac{π}{3}=\frac{s}{2s}=\frac{1}{2} \end{align*}\]

This result should not be surprising because, as we see from Figure \(\PageIndex{10}\), the side opposite the angle of \(\frac{π}{3}\) is also the side adjacent to \(\frac{π}{6}\), so \(\sin (\frac{π}{3})\) and \(\cos (\frac{π}{6})\) are exactly the same ratio of the same two sides, \(\sqrt{3} s\) and \(2s.\) Similarly, \( \cos (\frac{π}{3})\) and \( \sin (\frac{π}{6})\) are also the same ratio using the same two sides, \(s\) and \(2s\).

The interrelationship between the sines and cosines of \(\frac{π}{6}\) and \(\frac{π}{3}\) also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to π, π,and the right angle is \(\frac{π}{2}\), the remaining two angles must also add up to \(\frac{π}{2}\). That means that a right triangle can be formed with any two angles that add to \(\frac{π}{2}\)—in other words, any two complementary angles. So we may state a cofunction identity: If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure \(\PageIndex{11}\).

Using this identity, we can state without calculating, for instance, that the sine of \(\frac{π}{12}\) equals the cosine of \(\frac{5π}{12}\), and that the sine of \(\frac{5π}{12}\) equals the cosine of \(\frac{π}{12}\). We can also state that if, for a certain angle \(t, \cos t= \frac{5}{13},\) then \( \sin (\frac{π}{2}−t)=\frac{5}{13}\) as well.

Using Trigonometric Functions

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

Example \(\PageIndex{6}\): Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure \(\PageIndex{12}\).

Solution

We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

\[ \tan (30°)= \dfrac{7}{a} \nonumber\]

We rearrange to solve for \(a\).

\[\begin{align} a &=\dfrac{7}{ \tan (30°)} \\ & =12.1 \end{align} \nonumber\]

We can use the sine to find the hypotenuse.

\[ \sin (30°)= \dfrac{7}{c} \nonumber\]

Again, we rearrange to solve for \(c\).

\[\begin{align*} c &= \dfrac{7}{\sin (30°)} =14 \end{align*}\]

Using Right Triangle Trigonometry to Solve Applied Problems

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. See Figure \(\PageIndex{13}\).

Example \(\PageIndex{7}\): Measuring a Distance Indirectly

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of 57°between a line of sight to the top of the tree and the ground, as shown in Figure \(\PageIndex{14}\). Find the height of the tree.

Solution

We know that the angle of elevation is \(57°\) and the adjacent side is 30 ft long. The opposite side is the unknown height.

The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of \(57°\), letting \(h\) be the unknown height.

\[\begin{array}{cl} \tan θ = \dfrac{\text{opposite}}{\text{adjacent}} & \text{} \\ \tan (57°) = \dfrac{h}{30} & \text{Solve for }h. \\ h=30 \tan (57°) & \text{Multiply.} \\ h≈46.2 & \text{Use a calculator.} \end{array} \]

The tree is approximately 46 feet tall.

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Key Equations

Cofunction Identities

\[\begin{align*} \cos t &= \sin ( \frac{π}{2}−t) \\ \sin t &= \cos (\frac{π}{2}−t) \\ \tan t &= \cot (\frac{π}{2}−t) \\ \cot t &= \tan (\frac{π}{2}−t) \\ \sec t &= \csc (\frac{π}{2}−t) \\ \csc t &= \sec (\frac{π}{2}−t) \end{align*}\]

Key Concepts

• We can define trigonometric functions as ratios of the side lengths of a right triangle. See Example.
• The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle. See Example.
• We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur. See Example.
• Any two complementary angles could be the two acute angles of a right triangle.
• If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa. See Example.
• We can use trigonometric functions of an angle to find unknown side lengths.
• Select the trigonometric function representing the ratio of the unknown side to the known side. See Example.
• Right-triangle trigonometry permits the measurement of inaccessible heights and distances.
• The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known. See Example.

Glossary

adjacent side

in a right triangle, the side between a given angle and the right angle

angle of depression

the angle between the horizontal and the line from the object to the observer’s eye, assuming the object is positioned lower than the observer

angle of elevation

the angle between the horizontal and the line from the object to the observer’s eye, assuming the object is positioned higher than the observer

opposite side

in a right triangle, the side most distant from a given angle

hypotenuse

the side of a right triangle opposite the right angle