# 7.3: Right Triangle Trigonometry

Skills to Develop

- Use right triangles to evaluate trigonometric functions.
- 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.

Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. Measuring its height is no easy task and, in fact, the actual measurement has been a source of controversy for hundreds of years. The measurement process involves the use of triangles and a branch of mathematics known as trigonometry. In this section, we will define a new group of functions known as trigonometric functions, and find out how they can be used to measure heights, such as those of the tallest mountains.

### Using Right Triangles to Evaluate Trigonometric Functions

**Figure 7.3.1** shows a **right triangle** with a vertical side of lengthy y and a horizontal side has lengthx. x. Notice that the triangle is inscribed in a circle of radius 1. Such a circle, with a center at the origin and a radius of 1, is known as a **unit circle**.

**Figure 7.3.1**

We can define the trigonometric functions in terms an angle t and the lengths of the sides of the triangle. The **adjacent side** is the side closest to the angle, x. (Adjacent means “next to.”) The **opposite side** is the side across from the angle, y. The **hypotenuse** is the side of the triangle opposite the right angle, 1. These sides are labeled in **Figure 7.3.2**.

**Figure 7.3.2 **The sides of a right triangle in relation to angle \(t\)

Given a right triangle with an acute angle of \(t\) the first three trigonometric functions are listed.

Sine sin \(t\)=\(\dfrac{opposite}{hypotenuse}\)

Cosine cos \(t\)=\(\dfrac{adjacent}{hypotenuse}\)

Tangent tan \(t\)=\(\dfrac{opposite}{adjacent}\)

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.”

For the triangle shown in **Figure 7.3.1**, we have the following.

sin \(t\)=\(\dfrac{y}{1}\)

cos \(t\)=\(\dfrac{x}{1}\)

sec \(t\)=\(\dfrac{y}{x}\)

How to:Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle.

- Find the sine as the ratio of the opposite side to the hypotenuse.
- Find the cosine as the ratio of the adjacent side to the hypotenuse.
- Find the tangent as the ratio of the opposite side to the adjacent side.

Example

Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in **Figure 7.3.3**, find the value ofcosα. cos α.

**Figure 7.3.3**

**Solution:**

The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17.

\[cos(\alpha)=\dfrac{adjacent}{hypotenuse}\]

\[=\dfrac{15}{17}\]

Exercise 2.4.1

Given the triangle shown in **Figure 7.3.4**, find the value of sin \(t\).

**Figure 7.3.4**

**Solution:**

\(\dfrac{7}{25}\)

#### Reciprocal Functions

In addition to sine, cosine, and tangent, there are three more functions. These too are defined in terms of the sides of the triangle.

Secant sec \(t\)=\(\dfrac{hypotenuse}{adjacent}\)

Cosecant csc \(t\)=\(\dfrac{hypotenuse}{opposite}\)

Cotangent cot \(t\)=\(\dfrac{adjacent}{opposite}\)

Take another look at these definitions. These functions are the reciprocals of the first three functions.

\(sin\) \(t\)=\(\dfrac{1}{csc\space t}\) \(csc\) \(t\)=\(\dfrac{1}{sin\space t}\)

\(cos\space t=\dfrac{1}{sec\space t}\) \(sec\space t=\dfrac{1}{cos\space t}\)

\(tan\space t=\dfrac{1}{cot\space t}\) \(cot\space t=\dfrac{1}{tan\space t}\)

When working with right triangles, keep in mind that 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 7.3.5**. The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

**Figure 7.3.5 **The side adjacent to one angle is opposite the other angle.

Many problems ask for all six trigonometric functions for a given angle in a triangle. A possible strategy to use is to find the sine, cosine, and tangent of the angles first. Then, find the other trigonometric functions easily using the reciprocals.

How to:Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.

- If needed, draw the right triangle and label the angle provided.
- Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
- Find the required function:

- sine as the ratio of the opposite side to the hypotenuse
- cosine as the ratio of the adjacent side to the hypotenuse
- tangent as the ratio of the opposite side to the adjacent side
- secant as the ratio of the hypotenuse to the adjacent side
- cosecant as the ratio of the hypotenuse to the opposite side
- cotangent as the ratio of the adjacent side to the opposite side

Example

Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in **Figure 7.3.6**, evaluate sin \(α\),cos \(α\),tan \(α\),sec \(α\),csc \(α\),and cot \(α\).

**Figure 7.3.6**

**Solution:**

\[sin\space \alpha=\dfrac{opposite \alpha}{hypotenuse}=\dfrac{4}{5}\]

\[cos\space \alpha=\dfrac{adjacent to \alpha}{hypotenuse}=\dfrac{3}{5}\]

\[tan\space \alpha=\dfrac{opposite \alpha}{adjacent to \alpha}=\dfrac{4}{3}\]

\[sec\space \alpha=\dfrac{hypotenuse}{adjacent to \alpha}=\dfrac{5}{3}\]

\[csc\space \alpha=\dfrac{hypotenuse}{opposite \alpha}=\dfrac{5}{4}\]

\[cot\space \alpha=\dfrac{adjacent to \alpha}{opposite \alpha}=\dfrac{3}{4}\]

Analysis

Another approach would have been to find sine, cosine, and tangent first. Then find their reciprocals to determine the other functions.

\[sec\space \alpha=\dfrac{1}{cos\space \alpha}=\dfrac{1}{\dfrac{3}{5}}=\dfrac{5}{3}\]

\[csc\space \alpha=\dfrac{1}{sec\space \alpha}=\dfrac{1}{\dfrac{4}{5}}=\dfrac{5}{4}\]

\[cot\space \alpha=\dfrac{1}{tan\space \alpha}=\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\]

Exercise

Using the triangle shown in **Figure 7.3.7**, evaluate sin \(t\),cos \(t\), tan \(t\), sec \(t\), csc \(t\),and cot \(t\).

**Figure 7.3.7**

**Solution:**

sin \(t\)=\(\dfrac{33}{65}\), sec \(t\)=\(\dfrac{65}{56}\),

cos \(t\)=\(\dfrac{56}{65}\), csc \(t\)=\(\dfrac{65}{33}\),

tan \(t\)=\(\dfrac{33}{56}\), cot \(t\)=\(\dfrac{56}{33}\)

#### Finding Trigonometric Functions of Special Angles Using Side Lengths

It is helpful to evaluate the trigonometric functions as they relate to the special angles—multiples of30°,60°,and45°. Remember, however, that when dealing with right triangles, we are limited to angles between0° and 90°.

Suppose we have a 30°, 60°, 90° triangle, which can also be described as a \(\dfrac{\pi}{6}\), \(\dfrac{\pi}{3}\),\(\dfrac{\pi}{2}\) triangle. The sides have lengths in the relation \(s\), \(s\sqrt{3}\), \(2s\). The sides of a 45°,45°,90° triangle, which can also be described as a \(\dfrac{\pi}{4}\), \(\dfrac{\pi}{4}\),\(\dfrac{\pi}{2}\) triangle, have lengths in the relation \(s\), \(s\), \(\sqrt{2}s\). These relations are shown in **Figure 7.3.8**.

**Figure 7.3.8 **Side lengths of special triangles

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

How to:Given trigonometric functions of a special angle, evaluate using side lengths.

- Use the side lengths shown in
**Figure 7.3.8**for the special angle you wish to evaluate. - Use the ratio of side lengths appropriate to the function you wish to evaluate.

Example

Evaluating Trigonometric Functions of Special Angles Using Side Lengths

Find the exact value of the trigonometric functions of \(\dfrac{\pi}{3}\) using side lengths.

**Solution:**

\[sin(\dfrac{\pi}{3})=\dfrac{opp}{hyp}=\dfrac{\sqrt{3}s}{2s}=\dfrac{\sqrt{3}}{2}\]

\[cos(\dfrac{\pi}{3})=\dfrac{adj}{hyp}=\dfrac{s}{2s}=\dfrac{1}{2}\]

\[tan(\dfrac{\pi}{3})=\dfrac{opp}{adj}=\dfrac{\sqrt{3}s}{s}=\sqrt{3}\]

\[sec(\dfrac{\pi}{3})=\dfrac{hyp}{adj}=\dfrac{2s}{s}=2\]

\[csc(\dfrac{\pi}{3})=\dfrac{hyp}{opp}=\dfrac{2s}{\sqrt{3}s}=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\]

\[cot(\dfrac{\pi}{3})=\dfrac{adj}{opp}=\dfrac{s}{\sqrt{3}s}=\dfrac{\sqrt{3}}{3}\]

Exercise 2.4.1

Find the exact value of the trigonometric functions of \(\dfrac{\pi}{4}\),using side lengths.

**Solution:**

\(sin(\dfrac{\pi}{4})=\dfrac{\sqrt{2}}{2}\), \(cos(\dfrac{\pi}{4})=\dfrac{\sqrt{2}}{2}\), \(tan(\dfrac{\pi}{4})=1\)

\(sec(\dfrac{\pi}{4})=\sqrt{2}\), \(csc(\dfrac{\pi}{4})=\sqrt{2}\), \(cot(\dfrac{\pi}{4})=1\)

### Using Equal Cofunction of Complements

If we look more closely at the relationship between the sine and cosine of the special angles, we notice a pattern. In a right triangle with angles of \(\dfrac{\pi}{6}) and \(\dfrac{\pi}{3}\),we see that the sine of \(\dfrac{\pi}{3}\), ,namely \(\dfrac{\sqrt{3}}{2}\),is also the cosine of \(dfrac{\pi}{6}\),while the sine of \(\dfrac{\pi}{6}\),namely \(\dfrac{1}{2}\),is also the cosine of \(\dfrac{\pi}{3}\).

\[sin\dfrac{\pi}{3}=cos\dfrac{\pi}{6}=\dfrac{\sqrt{3}s}{2s}=\dfrac{\sqrt{3}}{2}\]

\[sin\dfrac{\pi}{6}=cos\dfrac{\pi}{3}=\dfrac{s}{2s}=\dfrac{1}{2}\]

See **Figure 7.3.9**

**Figure 7.3.9 **The sine ofπ3 π3 equals the cosine ofπ6 π6 and vice versa.

This result should not be surprising because, as we see from **Figure 7.3.9**, the side opposite the angle of \(\dfrac{\pi}{3}\) is also the side adjacent to \(\dfrac{\pi}{6}\),so \(sin(\dfrac{\pi}{3})\) and \(cos(\dfrac{\pi}{6})\) are exactly the same ratio of the same two sides, \(\sqrt{3}s\) and \(2s\). Similarly, \(cos(\dfrac{\pi}{3})\) and \(sin(\dfrac{\pi}{6})\) are also the same ratio using the same two sides, \(s\) and \(2s\).

The interrelationship between the sines and cosines of \(\dfrac{\pi}{6}\) and \(\dfrac{\pi}{3}\), 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 toπ, π,and the right angle is \(\dfrac{\pi}{2}\),the remaining two angles must also add up to \(\dfrac{\pi}{2}\). That means that a right triangle can be formed with any two angles that add to \(\dfrac{\pi}{2}\) —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 7.3.10**.

**Figure 7.3.10 **Cofunction identity of sine and cosine of complementary angles

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

A General Note: COFUNCTION IDENTITIES

The cofunction identities in radians are listed in **Table 7.3.1**.

\(cos\space t=sin\space (\dfrac{\pi}{2}−t)\) | \(sin\space t=cos\space (\dfrac{\pi}{2}−t)\) |

\(tan\space t=cot\space (\dfrac{\pi}{2}−t)\) | \(cot\space t=tan\space (\dfrac{\pi}{2}−t)\) |

\(sec\space t=csc\space (\dfrac{\pi}{2}−t)\) | \(csc\space t=sec\space (\dfrac{\pi}{2}−t)\) |

** Table 7.3.1**

How to:** Given the sine and cosine of an angle, find the sine or cosine of its complement.**

- To find the sine of the complementary angle, find the cosine of the original angle.
- To find the cosine of the complementary angle, find the sine of the original angle.

Example

Using Cofunction Identities

If \(sin\space t=\dfrac{5}{12}\), find \(cos\space (\dfrac{\pi}{2}−t\)).

**Solution:**

According to the cofunction identities for sine and cosine, we have the following.

\[sin\space t=cos\space (\dfrac{\pi}{2}−t)\]

So

\[cos\space(\dfrac{\pi}{2}−t)=\dfrac{5}{12}\]

Exercise

If \(csc\space (\dfrac{\pi}{6})=2\), find \(sec\space (\dfrac{\pi}{3}).\)

**Solution:**

2

### 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.

How to:** Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides.**

- For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
- Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
- Using the value of the trigonometric function and the known side length, solve for the missing side length.

Example

Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in **Figure 7.3.11**.

**Figure 7.3.11**

**Solution:**

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

\[tan\space (30°)=\dfrac{7}{a}\]

We rearrange to solve fora. a.

\[a=\dfrac{7}{tan(30°)}\]

\[≈12.1\]

We can use the sine to find the hypotenuse.

\[sin\space (30°)=\dfrac{7}{c}\]

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

\[c=\dfrac{7}{sin\space (30°)}\]

\[≈14\]

Exercise 2.4.1

A right triangle has one angle of \(\dfrac{\pi}{3}\) and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

**Solution:**

adjacent=\(10\);opposite=\(10\sqrt{3}\); missing angle is \(\dfrac{\pi}{6}\)

### 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 7.3.12**.

**Figure 7.3.12**

How to:** Given a tall object, measure its height indirectly.**

- Make a sketch of the problem situation to keep track of known and unknown information.
- Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
- At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
- Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
- Solve the equation for the unknown height.

Example

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 7.3.13**. Find the height of the tree.

**Figure 7.3.13**

**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.

\[tan\space \theta=\dfrac{opposite}{adjacent}\]

\(tan\space (57°)=\dfrac{h}{30}\) Solve for \(h\)

\(h=30tan\space (57°)\) Multiply

\(h≈46.2\) Use a calculator

The tree is approximately 46 feet tall.

Exercise

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of \(\dfrac{5\pi}{12}\) with the ground? Round to the nearest foot.

**Solution:**

About 52 ft

Media

Access these online resources for additional instruction and practice with right triangle trigonometry.

### 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. SeeExample.
- 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 facilitates 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.