Right Triangle Trigonometry

Learning Objectives

In this section you will:

• Use right triangles to evaluate trigonometric functions.
• Find function values for 30°(π6),45°(π4),30°(𝜋6),45°(𝜋4), and 60°(π3).60°(𝜋3).
• 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. 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 1 shows a right triangle with a vertical side of length y𝑦 and a horizontal side has length x.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 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 2.

Figure 2 The sides of a right triangle in relation to angle t𝑡

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

Sinesin t=oppositehypotenuseSinesin 𝑡=oppositehypotenuse

Cosinecos t=adjacenthypotenuseCosinecos 𝑡=adjacenthypotenuse

Tangenttan t=oppositeadjacentTangenttan 𝑡=oppositeadjacent

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “underlineSend underlineine is underlineoend underlinepposite over underlinehend underlineypotenuse, underlineCend underlineosine is underlineaend underlinedjacent over underlinehend underlineypotenuse, underlineTend underlineangent is underlineoend underlinepposite over underlineaend underlinedjacent.”

For the triangle shown in Figure 1, we have the following.

sin tcos ttan t===y1x1yxsin 𝑡=𝑦1cos 𝑡=𝑥1tan 𝑡=𝑦𝑥

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.

1. Find the sine as the ratio of the opposite side to the hypotenuse.
2. Find the cosine as the ratio of the adjacent side to the hypotenuse.
3. Find the tangent as the ratio of the opposite side to the adjacent side.

EXAMPLE 1

Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure 3, find the value of cosα.cos𝛼.

Figure 3

Solution

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

cos(α)==adjacenthypotenuse1517cos(𝛼)=adjacenthypotenuse=1517

TRY IT #1

Given the triangle shown in Figure 4, find the value of sint.sin𝑡.

Figure 4

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.

Secantsec t=hypotenuseadjacentSecantsec 𝑡=hypotenuseadjacent

Cosecantcsc t=hypotenuseoppositeCosecantcsc 𝑡=hypotenuseopposite

Cotangentcot t=adjacentoppositeCotangentcot 𝑡=adjacentopposite

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

sin tcos ttan t===1csc t1sec t1cot tcsc tsec tcot t===1sin t1cos t1tan tsin 𝑡=1csc 𝑡csc 𝑡=1sin 𝑡cos 𝑡=1sec 𝑡sec 𝑡=1cos 𝑡tan 𝑡=1cot 𝑡cot 𝑡=1tan 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 5. The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

Figure 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.

1. If needed, draw the right triangle and label the angle provided.
2. Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
3. 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 2

Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure 6, evaluate sinα,cosα,tanα,secα,cscα,andcotα.sin𝛼,cos𝛼,tan𝛼,sec𝛼,csc𝛼,andcot𝛼.

Figure 6

Solution

sin αcos αtan αsec αcsc αcot α=opposite αhypotenuse=adjacent to αhypotenuse=opposite αadjacent to α=hypotenuseadjacent to α=hypotenuseopposite α=adjacent to αopposite α=45=35=43=53=54=34sin 𝛼=opposite 𝛼hypotenuse=45cos 𝛼=adjacent to 𝛼hypotenuse=35tan 𝛼=opposite 𝛼adjacent to 𝛼=43sec 𝛼=hypotenuseadjacent to 𝛼=53csc 𝛼=hypotenuseopposite 𝛼=54cot 𝛼=adjacent to 𝛼opposite 𝛼=34

Analysis

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

sec α=1cos α=135=53sec 𝛼=1cos 𝛼=135=53

csc α=1sin α=145=54csc 𝛼=1sin 𝛼=145=54

cot α=1tan α=143=34cot 𝛼=1tan 𝛼=143=34

TRY IT #2

Using the triangle shown in Figure 7,evaluate sint,cost,tant,sect,csct,andcott.sin𝑡,cos𝑡,tan𝑡,sec𝑡,csc𝑡,andcot𝑡.

Figure 7

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 of 30°,60°,30°,60°, and 45°.45°. Remember, however, that when dealing with right triangles, we are limited to angles between 0° and 90°.0° and 90°.

Suppose we have a 30°,60°,90°30°,60°,90° triangle, which can also be described as a π6,π3,π2𝜋6,𝜋3,𝜋2 triangle. The sides have lengths in the relation s,3–√s,2s.𝑠,3𝑠,2𝑠. The sides of a 45°,45°,90°45°,45°,90° triangle, which can also be described as a π4,π4,π2𝜋4,𝜋4,𝜋2 triangle, have lengths in the relation s,s,2–√s.𝑠,𝑠,2𝑠. These relations are shown in Figure 8.

Figure 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.

1. Use the side lengths shown in Figure 8 for the special angle you wish to evaluate.
2. Use the ratio of side lengths appropriate to the function you wish to evaluate.

EXAMPLE 3

Evaluating Trigonometric Functions of Special Angles Using Side Lengths

Find the exact value of the trigonometric functions of π3,𝜋3, using side lengths.

Solution

sin(π3)cos(π3)tan(π3)sec(π3)csc(π3)cot(π3)=opphyp=adjhyp=oppadj=hypadj=hypopp=adjopp=3s√2s=s2s=3√ss=2ss=2s3√s=s3√s=3√2=12=3–√=2=23√=13√=23√3=3√3sin(𝜋3)=opphyp=3𝑠2𝑠=32cos(𝜋3)=adjhyp=𝑠2𝑠=12tan(𝜋3)=oppadj=3𝑠𝑠=3sec(𝜋3)=hypadj=2𝑠𝑠=2csc(𝜋3)=hypopp=2𝑠3𝑠=23=233cot(𝜋3)=adjopp=𝑠3𝑠=13=33

TRY IT #3

Find the exact value of the trigonometric functions of π4,𝜋4, using side lengths.

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 π6𝜋6 and π3,𝜋3, we see that the sine of π3,𝜋3, namely 3√2,32, is also the cosine of π6,𝜋6, while the sine of π6,𝜋6, namely 12,12, is also the cosine of π3.𝜋3.

sinπ3sinπ6=cosπ6=cosπ3=3√s2s=s2s=3√2=12sin𝜋3=cos𝜋6=3𝑠2𝑠=32sin𝜋6=cos𝜋3=𝑠2𝑠=12

See Figure 9.

Figure 9 The sine of π3𝜋3 equals the cosine of π6𝜋6 and vice versa.

This result should not be surprising because, as we see from Figure 9, the side opposite the angle of π3𝜋3 is also the side adjacent to π6,𝜋6, so sin(π3)sin(𝜋3) and cos(π6)cos(𝜋6) are exactly the same ratio of the same two sides, 3–√s3𝑠 and 2s.2𝑠. Similarly, cos(π3)cos(𝜋3) and sin(π6)sin(𝜋6) are also the same ratio using the same two sides, s𝑠 and 2s.2𝑠.

The interrelationship between the sines and cosines of π6𝜋6 and π3𝜋3 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 π2,𝜋2, the remaining two angles must also add up to π2.𝜋2. That means that a right triangle can be formed with any two angles that add to π2𝜋2 —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 10.

Figure 10 Cofunction identity of sine and cosine of complementary angles

Using this identity, we can state without calculating, for instance, that the sine of π12𝜋12 equals the cosine of 5π12,5𝜋12, and that the sine of 5π125𝜋12 equals the cosine of π12.𝜋12. We can also state that if, for a given angle t,cost=513,𝑡,cos𝑡=513, then sin(π2−t)=513sin(𝜋2−𝑡)=513 as well.

COFUNCTION IDENTITIES

The cofunction identities in radians are listed in Table 1.

cost=sin(π2−t)cos𝑡=sin(𝜋2−𝑡)	sint=cos(π2−t)sin𝑡=cos(𝜋2−𝑡)
tant=cot(π2−t)tan𝑡=cot(𝜋2−𝑡)	cott=tan(π2−t)cot𝑡=tan(𝜋2−𝑡)
sect=csc(π2−t)sec𝑡=csc(𝜋2−𝑡)	csct=sec(π2−t)csc𝑡=sec(𝜋2−𝑡)

HOW TO

Given the sine and cosine of an angle, find the sine or cosine of its complement.

1. To find the sine of the complementary angle, find the cosine of the original angle.
2. To find the cosine of the complementary angle, find the sine of the original angle.

EXAMPLE 4

Using Cofunction Identities

If sint=512,sin𝑡=512, find cos(π2−t).cos(𝜋2−𝑡).

Solution

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

sint=cos(π2−t)sin𝑡=cos(𝜋2−𝑡)

So

cos(π2−t)=512cos(𝜋2−𝑡)=512

TRY IT #4

If csc(π6)=2,csc(𝜋6)=2, find sec(π3).sec(𝜋3).

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.

1. 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.
2. Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
3. Using the value of the trigonometric function and the known side length, solve for the missing side length.

EXAMPLE 5

Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure 11.

Figure 11

Solution

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

tan(30°)=7atan(30°)=7𝑎

We rearrange to solve for a.𝑎.

a=≈7tan(30°)12.1𝑎=7tan(30°)≈12.1

We can use the sine to find the hypotenuse.

sin(30°)=7csin(30°)=7𝑐

Again, we rearrange to solve for c.𝑐.

c==7sin(30°)14𝑐=7sin(30°)=14

TRY IT #5

A right triangle has one angle of π3𝜋3 and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

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 12.

Figure 12

HOW TO

Given a tall object, measure its height indirectly.

1. Make a sketch of the problem situation to keep track of known and unknown information.
2. Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
3. 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.
4. Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
5. Solve the equation for the unknown height.

EXAMPLE 6

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°57° between a line of sight to the top of the tree and the ground, as shown in Figure 13. Find the height of the tree.

Figure 13

Solution

We know that the angle of elevation is 57°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°,57°, letting hℎ be the unknown height.

tan θtan(57°)hh===≈oppositeadjacenth3030tan(57°)46.2Solve for h.Multiply.Use a calculator.tan 𝜃=oppositeadjacenttan(57°)=ℎ30Solve for ℎ.ℎ=30tan(57°)Multiply.ℎ≈46.2Use a calculator.

The tree is approximately 46 feet tall.

TRY IT #6

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 5π125𝜋12 with the ground? Round to the nearest foot.