Using Right Triangles to Evaluate Trigonometric Functions

In earlier sections, we used a unit circle to define the trigonometric functions. In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of [latex]t[/latex] is its value at [latex]t[/latex] radians. First, we need to create our right triangle. Figure 1 shows a point on a unit circle of radius 1. If we drop a vertical line segment from the point [latex]\left(x,y\right)\\[/latex] to the x-axis, we have a right triangle whose vertical side has length [latex]y[/latex] and whose horizontal side has length [latex]x[/latex]. We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.

Figure 1

We know

Likewise, we know

These ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using [latex]\left(x,y\right)[/latex] coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of [latex]x[/latex], we will call the side between the given angle and the right angle the adjacent side to angle [latex]t[/latex]. (Adjacent means “next to.”) Instead of [latex]y[/latex], we will call the side most distant from the given angle the opposite side from angle [latex]t[/latex]. And instead of [latex]1[/latex], we will call the side of a right triangle opposite the right angle the hypotenuse. These sides are labeled in Figure 2.

Figure 2. The sides of a right triangle in relation to angle [latex]t[/latex].

Understanding Right Triangle Relationships

Given a right triangle with an acute angle of [latex]t[/latex],

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.”

Example 1: Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure 3, find the value of [latex]\cos \alpha[/latex].

Figure 3

Show Solution

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

[latex]\begin{align}\cos \left(\alpha \right)=\frac{\text{adjacent}}{\text{hypotenuse}} =\frac{15}{17} \end{align}[/latex]

Try It

Given the triangle shown in Figure 4, find the value of [latex]\text{sin}t[/latex].

Figure 4

Show Solution

[latex]\frac{7}{25}[/latex]

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 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.

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 2: Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure 6, evaluate [latex]\sin \alpha[/latex], [latex]\cos \alpha[/latex], [latex]\tan \alpha[/latex], [latex]\sec \alpha[/latex], [latex]\csc \alpha [/latex], and [latex]\cot \alpha[/latex].

Figure 6

Show Solution

[latex]\begin{align}&\sin \alpha =\frac{\text{opposite }\alpha }{\text{hypotenuse}}=\frac{4}{5}\\ &\cos \alpha =\frac{\text{adjacent to }\alpha }{\text{hypotenuse}}=\frac{3}{5} \\ &\tan \alpha =\frac{\text{opposite }\alpha }{\text{adjacent to }\alpha }=\frac{4}{3} \\ &\sec \alpha =\frac{\text{hypotenuse}}{\text{adjacent to }\alpha }=\frac{5}{3} \\ &\csc \alpha =\frac{\text{hypotenuse}}{\text{opposite }\alpha }=\frac{5}{4} \\ &\cot \alpha =\frac{\text{adjacent to }\alpha }{\text{opposite }\alpha }=\frac{3}{4} \end{align}[/latex]

Try It

Using the triangle shown in Figure 7, evaluate [latex]\sin t[/latex], [latex]\cos t[/latex], [latex]\tan t[/latex], [latex]\sec t[/latex], [latex]\csc t[/latex], and [latex]\cot t[/latex].

Figure 7

Show Solution

[latex]\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}[/latex]

Finding Trigonometric Functions of Special Angles Using Side Lengths

We have already discussed the trigonometric functions as they relate to the special angles on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special 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. We will use multiples of [latex]30^\circ [/latex], [latex]60^\circ [/latex], and [latex]45^\circ[/latex], however, remember that when dealing with right triangles, we are limited to angles between [latex]0^\circ \text{ and } 90^\circ[/latex].

Suppose we have a [latex]30^\circ ,60^\circ ,90^\circ [/latex] triangle, which can also be described as a [latex]\frac{\pi }{6}, \frac{\pi }{3},\frac{\pi }{2}[/latex] triangle. The sides have lengths in the relation [latex]s,\sqrt{3}s,2s[/latex]. The sides of a [latex]45^\circ ,45^\circ ,90^\circ [/latex] triangle, which can also be described as a [latex]\frac{\pi }{4},\frac{\pi }{4},\frac{\pi }{2}[/latex] triangle, have lengths in the relation [latex]s,s,\sqrt{2}s[/latex]. 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.

By looking at all six trig functions and their special angles, we can organize it all in one place.

Now that we have this table, we can use it to find the exact values of trigonometric expressions.

Using Equal Cofunction of Complements

If we look at the table above, we will notice a pattern. In a right triangle with angles of [latex]\frac{\pi }{6}[/latex] and [latex]\frac{\pi }{3}[/latex], we see that the sine of [latex]\frac{\pi }{3}[/latex], namely [latex]\frac{\sqrt{3}}{2}[/latex], is also the cosine of [latex]\frac{\pi }{6}[/latex], while the sine of [latex]\frac{\pi }{6}[/latex], namely [latex]\frac{1}{2}[/latex], is also the cosine of [latex]\frac{\pi }{3}[/latex].

This result should not be surprising because, as we see from Figure 9, the side opposite the angle of [latex]\frac{\pi }{3}[/latex] is also the side adjacent to [latex]\frac{\pi }{6}[/latex], so [latex]\sin \left(\frac{\pi }{3}\right)[/latex] and [latex]\cos \left(\frac{\pi }{6}\right)[/latex] are exactly the same ratio of the same two sides, [latex]\sqrt{3}s[/latex] and [latex]2s[/latex]. Similarly, [latex]\cos \left(\frac{\pi }{3}\right)[/latex] and [latex]\sin \left(\frac{\pi }{6}\right)[/latex] are also the same ratio using the same two sides, [latex]s[/latex] and [latex]2s[/latex].

The interrelationship between the sines and cosines of [latex]\frac{\pi }{6}[/latex] and [latex]\frac{\pi }{3}[/latex] 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 [latex]\pi [/latex], and the right angle is [latex]\frac{\pi }{2}[/latex], the remaining two angles must also add up to [latex]\frac{\pi }{2}[/latex]. That means that a right triangle can be formed with any two angles that add to [latex]\frac{\pi }{2}[/latex] —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 [latex]\frac{\pi }{12}[/latex] equals the cosine of [latex]\frac{5\pi }{12}[/latex], and that the sine of [latex]\frac{5\pi }{12}[/latex] equals the cosine of [latex]\frac{\pi }{12}[/latex]. We can also state that if, for a certain angle [latex]t[/latex], [latex]\cos \text{ }t=\frac{5}{13}[/latex], then [latex]\sin \left(\frac{\pi }{2}-t\right)=\frac{5}{13}[/latex] 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 7: Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure 11.

Figure 11

Show Solution

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

[latex]\tan \left(30^\circ \right)=\frac{7}{a}[/latex]

We rearrange to solve for [latex]a[/latex].

[latex]\begin{align}a&=\frac{7}{\tan \left(30^\circ \right)} \\ &=12.1\end{align}[/latex]

We can use the sine to find the hypotenuse.

[latex]\sin \left(30^\circ \right)=\frac{7}{c}[/latex]

Again, we rearrange to solve for [latex]c[/latex].

[latex]\begin{align}c&=\frac{7}{\sin \left(30^\circ \right)} \\ &=14 \end{align}[/latex]

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.  In the figure below, [latex]\alpha[/latex] represents the angle of elevation and [latex]\beta[/latex] represents the angle of depression.

Figure 12

Example 8: 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 the angle of elevation to be [latex]57^\circ [/latex], as shown in Figure 13. Find the height of the tree.

Figure 13

Show Solution

We know that the angle of elevation is [latex]57^\circ [/latex] 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 [latex]57^\circ [/latex], letting [latex]h[/latex] be the unknown height.

[latex]\begin{align}&\tan \theta =\frac{\text{opposite}}{\text{adjacent}} \\ &\tan\left(57^\circ \right)=\frac{h}{30}&& \text{Solve for }h. \\ &h=30\tan \left(57^\circ \right)&& \text{Multiply}.\\ &h\approx 46.2&& \text{Use a calculator}. \end{align}[/latex]

The tree is approximately 46 feet tall.

Inverse Trigonometric Function on a calculator

If two sides of a right triangle are given, an inverse trigonometric function can be used to find an acute angle in a triangle. There is an inverse trig button on a calculator that appears above the normal sin, cos, tan buttons on a calculator. A shift or second key is required to select these, and they are [latex]\sin^{-1},\cos^{-1},\tan^{-1}[/latex]. The calculator will return an angle in either radians or degrees, depending on what mode your calculator is in.  We will need the inverse tangent function for bearing.

Bearing

Bearing is the direction you are heading according to a compass. On most maps, N is up, S is down, W is to the left, and E is to the right. Bearings are written in this form: (N or S)(Acute Angle)(E or W). The angle is measured from either the north or south, whichever letter comes first in your bearing. For example, [latex]N20^\circ E[/latex] means that you will go north and then go 20 degrees to the east, or to the right. The drawing below gives an example of how bearings are drawn. As you can see, each angle is measure from either the N or S depending on the first letter of the bearing. Bearings are NOT drawn in standard position, which means the are NOT drawn from the positive x-axis.

Now we will will look at some application problems that involve bearings and right triangles.

Example 9: Find the bearing

A semi travels east for 8 miles, makes a right turn, and then travels south for 11 more miles. What is the bearing from the starting point to the semi’s current location?

Show Solution

First we need to draw a picture of the problem:

We want to find the angle [latex]\alpha[/latex] that is inside the triangle. To do this, we will set up a trig definition that relates the two given sides. We will choose tangent. Then we will take the inverse tangent to find the angle:
[latex]\begin{align} &\tan\left(\alpha\right)=\frac{11}{8} \\& \tan^{-1}\left(\tan\left(\alpha\right)\right)=\tan^{-1}\left(\frac{11}{8}\right) \\ &\alpha=\tan^{-1}\left(\frac{11}{8}\right) \\& \alpha\approx 54^\circ \end{align}[/latex]

The bearing must be measured from south since the triangle is drawn in the fourth quadrant. On the picture, we must find [latex]\beta[/latex]. Since [latex]\alpha+\beta=90^\circ[/latex], then [latex]\beta=90^\circ-\alpha=36^\circ[/latex]. Therefore our bearing is [latex]S 36^\circ E[/latex].

Example 10: Find the bearing

A jeep travels on a bearing of [latex]N 40^\circ W[/latex] for 6 miles. How far north and how far west is the jeep from its starting point? Round your answers to one decimal place.

Show Solution

First we need to draw a picture of the problem:

We need to find the angle [latex]\alpha[/latex] inside of the triangle before we set up a trig equation: [latex]\alpha=90^\circ-40^\circ=50^\circ[/latex].

Next we will set up a trig function to find the indicated sides of the triangle:

First we will find the number of miles traveled North:

[latex]\begin{align} & \sin\left(50^\circ\right)=\frac{N}{6} \\& 6\sin\left(50^\circ\right)=N \\ & N\approx 4.6\text{ miles} \end{align}[/latex]

Now we will find the number of miles traveled west:

[latex]\begin{align} & \cos\left(50^\circ\right)=\frac{W}{6} \\& 6\cos\left(50^\circ\right)=W \\ & W\approx 3.9\text{ miles} \end{align}[/latex]

Convert Percent Grade Into Degrees

Percent grade is often seen on roadways or trails, and it is a measure of steepness.  For example, if a road has a 6% grade, this means the road rises 6 feet over a horizontal distance (run) of 100 feet.  Percent grade is found by dividing the rise over the run as shown in Figure 14. The rise and run are also the opposite and adjacent sides.  We can find [latex]\theta[/latex] by taking the inverse tangent.

Figure 14

In order to calculate the degree measurement of a percent grade, first change the percent into a decimal by dividing by 100. Then take the inverse tangent of this decimal, and this will give the angle in degrees.  This is actually the angle of elevation, which we studied earlier in this section!

Key Equations

Section 4.3 Homework Exercises

1. For the given right triangle, label the adjacent side, opposite side, and hypotenuse for the indicated angle.

2. When a right triangle with a hypotenuse of 1 is placed in the unit circle, which sides of the triangle correspond to the x- and y-coordinates?

3. The tangent of an angle compares which sides of the right triangle?

4. What is the relationship between the two acute angles in a right triangle?

5. Explain the cofunction identity.

For the following exercises, evaluate the expression.  Rationalize the denominator if necessary.

6. [latex]\sin\frac{\pi}{4}\cos\frac{\pi}{4}-\tan\frac{\pi}{3}[/latex]

7. [latex]\sin\frac{\pi}{4}\cos\frac{\pi}{3}-\tan\frac{\pi}{4}[/latex]

8. [latex]2\sec^{2}\frac{\pi}{4}+\cot^{2}\frac{\pi}{3}[/latex]

9. [latex]3\csc^{2}\frac{\pi}{3}+\cot^{2}\frac{\pi}{4}[/latex]

For the following exercises, use cofunctions of complementary angles.

10. [latex]\cos \left(34^\circ\right)=\sin \left(\text{__}^\circ\right)[/latex]

11. [latex]\cos \left(\frac{\pi }{3}\right)=\sin \text{(___)}[/latex]

12. [latex]\csc \left(21^\circ\right)=\sec \left(\text{___}^\circ \right)[/latex]

13. [latex]\tan \left(\frac{\pi }{4}\right)=\cot \left(\text{__}\right)[/latex]

For the following exercises, find the lengths of the missing sides if side [latex]a[/latex] is opposite angle [latex]A[/latex], side [latex]b[/latex] is opposite angle [latex]B[/latex], and side [latex]c[/latex] is the hypotenuse.

14. [latex]\cos B=\frac{4}{5},a=10[/latex]

15. [latex]\sin B=\frac{1}{2}, a=20[/latex]

16. [latex]\tan A=\frac{5}{12},b=6[/latex]

17. [latex]\tan A=100,b=100[/latex]

18. [latex]\sin B=\frac{1}{\sqrt{3}}, a=2[/latex]

19. [latex]a=5,\measuredangle A={60}^{\circ }[/latex]

20. [latex]c=12,\measuredangle A={45}^{\circ }[/latex]

For the following exercises, use Figure 14 to evaluate each trigonometric function of angle [latex]A[/latex].

Figure 14

21. [latex]\sin A[/latex]

22. [latex]\cos A[/latex]

23. [latex]\tan A[/latex]

24. [latex]\csc A[/latex]

25. [latex]\sec A[/latex]

26. [latex]\cot A[/latex]

For the following exercises, use Figure 15 to evaluate each trigonometric function of angle [latex]A[/latex].

Figure 15

27. [latex]\sin A[/latex]

28. [latex]\cos A[/latex]

29. [latex]\tan A[/latex]

30. [latex]\csc A[/latex]

31. [latex]\sec A[/latex]

32. [latex]\cot A[/latex]

For the following exercises, solve for the unknown sides of the given triangle.

33.

34.

35.

For the following exercises, use a calculator to find the length of each side to four decimal places.

36.

37.

38.

39.

40.

41. [latex]b=15,\measuredangle B={15}^{\circ }[/latex]

42. [latex]c=200,\measuredangle B={5}^{\circ }[/latex]

43. [latex]c=50,\measuredangle B={21}^{\circ }[/latex]

44. [latex]a=30,\measuredangle A={27}^{\circ }[/latex]

45. [latex]b=3.5,\measuredangle A={78}^{\circ }[/latex]

46. Find [latex]x[/latex].

47. Find [latex]x[/latex].

48. Find [latex]x[/latex].

49. Find [latex]x[/latex].

50. A radio tower is located 400 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is [latex]36^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]23^\circ [/latex]. How tall is the tower?

51. A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is [latex]43^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]31^\circ [/latex]. How tall is the tower?

52. A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is [latex]15^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]2^\circ [/latex]. How far is the person from the monument?

53. A 400-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is [latex]18^\circ [/latex], and that the angle of depression to the bottom of the tower is [latex]3^\circ [/latex]. How far is the person from the monument?

54. There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be [latex]40^\circ [/latex]. From the same location, the angle of elevation to the top of the antenna is measured to be [latex]43^\circ [/latex]. Find the height of the antenna.

55. There is lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be [latex]36^\circ [/latex]. From the same location, the angle of elevation to the top of the lightning rod is measured to be [latex]38^\circ [/latex]. Find the height of the lightning rod.

56. A 33-ft ladder leans against a building so that the angle between the ground and the ladder is [latex]80^\circ [/latex]. How high does the ladder reach up the side of the building?

57. A 23-ft ladder leans against a building so that the angle between the ground and the ladder is [latex]80^\circ [/latex]. How high does the ladder reach up the side of the building?

58. The angle of elevation to the top of a building in New York is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building.

59. The angle of elevation to the top of a building in Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building.

60. Assuming that a 370-foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be [latex]60^\circ [/latex], how far from the base of the tree am I?

61. A car travels west for 5 miles, turns left, and then travels south for 9 miles.  What is the bearing from car’s starting position to its current position?  Round your answer to two decimal places.

62. A truck travels east for 4 miles, turns left, and then travels north for 6 miles.  What is the bearing from the truck’s starting position to its current position? Round your answer to two decimal places.

63. An ant travels on a bearing of [latex]N 22^\circ E[/latex] for 36 inches.  How far east and how far north is the ant from its starting position?  Round your answers to two decimal places.

64. A spider crawls on a bearing of [latex]S 34^\circ W[/latex] for 20 inches.  How far west and how far south is the spider from its starting point? Round your answers to two decimal places.

65. Originally built in 1901 by Colonel J.W. Eddy, Angels Flight in Los Angeles is said to be the world’s shortest incorporated railway.  The counterbalanced cars, controlled by cables, travel a 33% grade for 315 feet.  What is the angle the track makes with a horizontal line, rounded to one decimal place?

66. The Saluda Grade is the steepest standard-gauge mainline railway grade in the United States.  Between Melrose and Saluda, North Carolina, the maximum grade is 4.9% for about 300 feet.  What is the angle the track makes with a horizontal line, rounded to one decimal place?