right triangles and trigonometry homework 5

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Chapter 2: Trigonometric Ratios

2.3 Solving Right Triangles

Algebra refresher.

1. [latex]\:\sqrt{2}\sqrt{2}[/latex]

2. [latex]\:\dfrac{3}{\sqrt{3}}[/latex]

3. [latex]\:\sqrt{8}[/latex]

4. [latex]\:\sqrt{\dfrac{3}{4}}[/latex]

Rationalize the denominator.

5. [latex]\:\dfrac{1}{\sqrt{2}}[/latex]

6. [latex]\:\dfrac{2}{\sqrt{3}}[/latex]

7. [latex]\:\dfrac{6}{\sqrt{3}}[/latex]

8. [latex]\:\dfrac{4}{\sqrt{8}}[/latex]

[latex]\underline{\qquad\qquad\qquad\qquad}[/latex]

• [latex]\displaystyle 2\vphantom{\dfrac{\sqrt{3}}{2}}[/latex]
• [latex]\displaystyle \sqrt{3}\vphantom{\vphantom{\dfrac{\sqrt{3}}{2}}}[/latex]
• [latex]\displaystyle 2\sqrt{2}\vphantom{\dfrac{\sqrt{3}}{2}}[/latex]
• [latex]\displaystyle \dfrac{\sqrt{3}}{2}[/latex]
• [latex]\displaystyle \dfrac{\sqrt{2}}{2}[/latex]
• [latex]\displaystyle \dfrac{2\sqrt{3}}{3}[/latex]
• [latex]\displaystyle 2\sqrt{3}\vphantom{\dfrac{\sqrt{3}}{2}}[/latex]
• [latex]\displaystyle \sqrt{2}\vphantom{\dfrac{\sqrt{3}}{2}}[/latex]

Learning Objectives

• Solve a right triangle
• Use inverse trig ratio notation
• Use trig ratios to find an angle
• Solve problems involving right triangles
• Know the trig ratios for the special angles

Introduction

A triangle has six parts: three sides and three angles. In a right triangle, we know that one of the angles is [latex]90 {^o}\text{.}[/latex] If we know three parts of a right triangle, including one of the sides, we can use trigonometry to find all the other unknown parts. This is called solving the triangle.

Example 2.31.

The hypotenuse of a right triangle is 150 feet long, and one of the angles is [latex]35 {^o}\text{,}[/latex] as shown in the figure. Solve the triangle.

We can find the side opposite the 35° angle by using the sine ratio. [latex]\begin{align*} \sin 35 {^o} = \dfrac {\text{opposite}}{\text{hypotenuse}}\\ 0.5736 = \dfrac{a}{150}\\ a = 150(0.5736) = 86.04 \end{align*}[/latex] The opposite side is about 86 feet long. To find side [latex]b[/latex], we could use the Pythagorean theorem now, but it is better to use given information, rather than values we have calculated, to find the other unknown parts. We will use the cosine ratio. [latex]\begin{align*} \cos 35 {^o} = \dfrac {\text{adjacent}}{\text{hypotenuse}}\\ 0.8192 = \dfrac{b}{150}\\ b = 150(0.8192) = 122.89 \end{align*}[/latex] The adjacent side is about [latex]123[/latex] feet long. Finally, the unknown angle is the complement of [latex]35 {^o}\text{,}[/latex] or [latex]55 {^o}\text{.}[/latex]

Checkpoint 2.32.

Sketch a right triangle with

• one angle of [latex]37 {^o}\text{,}[/latex] and
• the side adjacent to that angle of length 5 centimeters.

Without doing the calculations, list the steps you would use to solve the triangle.

Use tan [latex]37{^o}[/latex] to find the opposite side. Use cos [latex]37{^o}[/latex] to find the hypotenuse. Subtract [latex]37{^o}[/latex] from [latex]90{^o}[/latex] to find the third angle.

Finding an Angle

While watching her niece at the playground, Francine wonders how steep the slide is. She happens to have a tape measure and her calculator with her and finds that the slide is 77 inches high and covers a horizontal distance of 136 inches, as shown below.

Francine knows that one way to describe the steepness of an incline is to calculate its slope, which in this case is [latex]\begin{equation*} \dfrac {\Delta y}{\Delta x} = \dfrac{77}{136} = 0.5662 \end{equation*}[/latex] However, Francine would really like to know what angle the slide makes with the horizontal. She realizes that the slope she has just calculated is also the tangent of the angle she wants.

If we know the tangent of an angle, can we find the angle? Yes, we can: locate the key labeled [latex]\boxed{\text{TAN}^{-1}}[/latex] on your calculator; it is probably the second function above the TAN key. Enter [latex]\qquad\qquad\qquad[/latex] 2nd TAN 0.5662

and you should find that [latex]\begin{equation*} \text{tan}^{-1} 0.5662 = 29.52 {^o}\text{.} \end{equation*}[/latex] This means that [latex]29.52 {^o}[/latex] is the angle whose tangent is [latex]0.5662\text{.}[/latex] We read the notation as “ inverse tangent of [latex]0.5662[/latex] is [latex]29.52[/latex] degrees.”

When we find [latex]\tan^{-1}[/latex] of a number, we are finding an angle whose tangent is that number. Similarly, [latex]\sin^{-1}[/latex] and [latex]cos^{-1}[/latex] are read as “inverse sine” and “inverse cosine.” They find an angle with the given sine or cosine.

Example 2.33.

Find the angle whose sine is [latex]0.6834\text{.}[/latex]

Enter 2nd SIN 0.6834 into your calculator to find [latex]\begin{equation*} \sin^{-1} 0.6834 = 43.11{^o} \end{equation*}[/latex] So [latex]43.11{^o}[/latex] is the angle whose sine is [latex]0.6834\text{.}[/latex] Or we can say that [latex]\begin{equation*} \sin 43.11{^o} = 0.6834 \end{equation*}[/latex] You can check the last equation on your calculator.

In the last example, the two equations:

[latex]\begin{equation*} \sin 43.11{^o} = 0.6834\end{equation*}[/latex]

[latex]\begin{equation*} \sin^{-1} 0.6834 = 43.11{^o}\end{equation*}[/latex]

say the same thing in different ways.

Caution 2.35.

The notation [latex]\sin^{-1} x[/latex] does not mean [latex]\dfrac{1}{\sin x}\text{.}[/latex] It is true that we use negative exponents to indicate reciprocals of numbers; for example, [latex]a^{-1} = \dfrac{1}{a}[/latex] and [latex]3^{-1} = \dfrac{1}{3}\text{.}[/latex] But “sin” by itself is not a variable.

• [latex]\sin^{-1} x[/latex] means “the angle whose sine is [latex]x[/latex]”
• [latex]\dfrac{1}{\sin x}[/latex] means “the reciprocal of the sine of angle [latex]x[/latex]”

(You may recall that [latex]f^{-1}(x)[/latex] denotes the inverse function for [latex]f(x)\text{.}[/latex] We will study trigonometric functions in Chapter 4.)

Checkpoint 2.36.

Write the following fact in two different ways: [latex]68{^o}[/latex] is the angle whose cosine is [latex]0.3746\text{.}[/latex]

[latex]\cos 68{^o} = 0.3746[/latex] or [latex]\cos^{-1}(0.3746) = 68{^o}[/latex]

Example 2.37.

Find the angle of inclination of a hill if you gain [latex]400[/latex] feet in elevation while traveling half a mile.

A sketch of the hill is shown at right. (Recall that [latex]1[/latex] mile [latex]= 5280[/latex] feet.)

[latex]\begin{align*} \sin \theta = \dfrac {400}{2640} = 0.\overline{15}\\ \theta = \sin^{-1} 0.\overline{15} = 8.71{^o} \end{align*}[/latex]

The angle of inclination of the hill is about [latex]8.7{^o}\text{.}[/latex]

Checkpoint 2.38.

The tallest living tree is a coastal redwood named Hyperion, at 378.1 feet tall. If you stand 100 feet from the base of the tree, what is the angle of elevation of your line of sight to the top of the tree? Round your answer to the nearest degree.

[latex]75{^o}[/latex]

The Special Angles

The trigonometric ratios for most angles are irrational numbers, but there are a few angles whose trig ratios are “nice” values. You already know one of these values: the sine of [latex]30{^o}\text{.}[/latex] Because the sides of a right triangle are related by the Pythagorean theorem, if we know any one of the trig ratios for an angle, we can find the others. Recall that the side opposite a [latex]30{^o}[/latex] angle is half the length of the hypotenuse, so [latex]\sin 30{^o} = \dfrac{1}{2}\text{.}[/latex]

The figure at right shows a 30-60-90 triangle with hypotenuse of length [latex]2[/latex]. The opposite side has length 1, and we can calculate the length of the adjacent side. [latex]\begin{align*} 1^2 + b^2 = 2^2 \\ b^2 = 2^2 - 1^2 = 3\\ b = \sqrt{3} \end{align*}[/latex]

Now we know the cosine and tangent of [latex]30{^o}\text{.}[/latex] [latex]\begin{equation*} \cos 30{^o} = \dfrac {\text{adjacent}}{\text{hypotenuse}} = \dfrac{\sqrt{3}}{2} \qquad \tan 30{^o} = \dfrac {\text{opposite}}{\text{adjacent}} = \dfrac{1}{\sqrt{3}} \end{equation*}[/latex]

These are exact values for the trig ratios, but we can also find decimal approximations. Use your calculator to verify the following approximate values. [latex]\begin{align*} \ \ {\text{exact value}} \ \ {\text{approximation}}\\ \cos 30{^o} = \dfrac{\sqrt{3}}{2} \approx 0.8660\\ \tan 30{^o} = \dfrac{1}{\sqrt{3}} \approx 0.5774 \end{align*}[/latex]

Caution 2.39.

It is important for you to understand the difference between exact and approximate values. These decimal approximations, like nearly all the other trig values your calculator gives you, are rounded off. Even if your calculator shows you ten or twelve digits, the values are not exactly correct — although they are quite adequate for most practical calculations!

The angles [latex]30{^o}\text{,}[/latex] [latex]60{^o}\text{,}[/latex] and [latex]45{^o}[/latex] are “special” because we can easily find exact values for their trig ratios and use those exact values to find exact lengths for the sides of triangles with those angles.

Example 2.40.

The sides of an equilateral triangle are [latex]8[/latex] centimeters long. Find the exact length of the triangle’s altitude, [latex]h\text{.}[/latex]

The altitude divides the triangle into two 30-60-90 right triangles as shown in the figure. The altitude is adjacent to the [latex]30{^o}[/latex] angle, and the hypotenuse of the right triangle is 8 centimeters. Thus,

[latex]\begin{align*} \cos 35 {^o} = \dfrac {\text{adjacent}}{\text{hypotenuse}} \ \ {\text{Fill in the values.}}\\ \dfrac{\sqrt{3}}{2} = \dfrac{h}{8} \ \ {\text{Multiply both sides by 8.}}\\ h = 8\left(\dfrac{\sqrt{3}}{2}\right) = 4\sqrt{3} \end{align*}[/latex] The altitude is exactly [latex]4\sqrt{3}[/latex] centimeters long.

From this exact answer, we can find approximations to any degree of accuracy we like. You can check that [latex]4\sqrt{3} \approx6.9282\text{,}[/latex] so the altitude is approximately 6.9 centimeters long.

Checkpoint 2.41.

Use the figure in the previous example to find exact values for the sine, cosine, and tangent of [latex]60{^o}\text{.}[/latex]

[latex]\sin 60{^o} = \dfrac{\sqrt{3}}{2}\text{,}[/latex] [latex]\cos 60{^o} = \dfrac{1}{2}\text{,}[/latex] [latex]\tan 60{^o} = \sqrt{3}[/latex]

There is one more special angle: [latex]45{^o}\text{.}[/latex] We find the trig ratios for this angle using an isosceles right triangle. Because the base angles of an isosceles triangle are equal, they must both be [latex]45{^o}\text{.}[/latex] The figure shows an isosceles right triangle with equal sides of length 1. You can use the Pythagorean theorem to show that the hypotenuse has length [latex]\sqrt{2}\text{,}[/latex] so the trig ratios for [latex]45{^o}[/latex] are[latex]\begin{align*} \sin 45{^o} = \dfrac {\text{opposite}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{2}} \approx 0.7071\\ \cos 45{^o} = \dfrac {\text{adjacent}}{\text{hypotenuse}} = \dfrac{1}{\sqrt{2}} \approx 0.7071\\ \tan 45{^o} = \dfrac {\text{opposite}}{\text{adjacent}} = 1 \end{align*}[/latex]

The Trigonometric Ratios for the Special Angles

Here is a summary of the trig ratios for the special angles.

You should memorize the exact values for these trig ratios. A good way to remember them is to know the two special triangles shown below. From these triangles, you can always write down the three trig ratios for the special angles.

You should also be able to recognize their decimal approximations.

We can use the special angles as benchmarks for estimating and mental calculation. For example, we know that [latex]60{^o} \approx 0.8660\text{,}[/latex] so if sin [latex]\theta = 0.95[/latex] for some unknown angle [latex]\theta\text{,}[/latex] we know that [latex]\theta \gt 60{^o}\text{,}[/latex] because as [latex]\theta[/latex] increases from the sine of [latex]\theta[/latex] increases also.

Example 2.43.

If [latex]\cos \alpha \gt \dfrac{\sqrt{3}}{2}\text{,}[/latex] what can we say about [latex]\alpha\text{?}[/latex]

As an angle increases from [latex]0{^o}[/latex] to [latex]90{^o}\text{,}[/latex] its cosine decreases. Now, [latex]\cos 30{^o} = \dfrac{\sqrt{3}}{2}\text{,}[/latex] so if [latex]\cos \alpha \gt \dfrac{\sqrt{3}}{2}\text{,}[/latex] then [latex]\alpha[/latex] must be less than [latex]30{^o}\text{.}[/latex]

Checkpoint 2.44.

If [latex]1 \lt \tan \beta \lt \sqrt{3}\text{,}[/latex] what can we say about [latex]\beta\text{?}[/latex]

[latex]45{^o} \lt \beta \lt 60{^o}[/latex]

Section 2.3 Summary

• Solve a triangle
• Inverse sine
• Inverse cosine
• Inverse tangent
• Special angles
• Exact value
• Decimal approximation
• If we know one of the sides of a right triangle and any one of the other four parts, we can use trigonometry to find all the other unknown parts.
• If we know one of the trigonometric ratios of an acute angle, we can find the angle using the inverse trig key on a calculator.
• For the trigonometric ratios of most angles, your calculator gives approximations, not exact values.

Study Questions

• How many parts of a right triangle (including the right angle) do you need to know in order to solve the triangle?
• Why is it better to use the given values, rather than values you have calculated, when solving a triangle?
• What is the [latex]\sin^{-1}[/latex] (or [latex]\cos^{-1}[/latex] or [latex]\tan^{-1}[/latex]) button on the calculator used for?
• Which are the “special” angles, and why are they special?

If we know three parts of a right triangle, including one of the sides, we can use trigonometry to find all the other unknown parts. This is called solving the triangle.

The inverse sine (arcsine) of a value is the angle whose sine is equal to that value.

The inverse cosine (arccosine) of a value is the angle whose sine is equal to that value.

The inverse tangent (arctangent) of a value is the angle whose sine is equal to that value.

Special angles in trigonometry are angles whose sine, cosine, and tangent values can be calculated exactly without the use of a calculator.

An exact value for a trigonometric function is a value that can be represented as a finite combination of integers, radicals, and/or known mathematical constants without the use of a calculator or approximations.

A decimal approximation for a trigonometric function is an estimation of its value expressed as a finite string of decimal digits, usually rounded to a certain number of decimal places, and obtained through the use of a calculator or other computational tool.

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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5.2: Solution of Right Triangles

• Last updated
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• Page ID 34143

• Henry Africk
• CUNY New York City College of Technology via New York City College of Technology at CUNY Academic Works

In the preceding section we showed that all \(30^{\circ}\) angles have the same trigonometric values. If we compute each of these values to four decimal places, we obtain \(\sin 30^{\circ} = \dfrac{1}{2} = 0.5000\), \(\cos 30^{\circ} = \dfrac{\sqrt{3}}{2} = \dfrac{1.73205}{2} = 0.8660\), and \(\tan 30^{\circ} = \dfrac{\sqrt{3}}{3} = \dfrac{1.73205}{3} = 0.5774\). These numbers appear in the table of trigonometric values on page 356 in the row corresponding to \(30^{\circ}\).

As you can see, this table contains the trigonometric values of angles from \(1^{\circ}\) to \(90^{\circ}.\) It is impractical to compute most of these values directly, so we will use this table when we need them. A pocket calculator with trigonometric functions may also be used.

Example \(\PageIndex{1}\)

Find \(\sin 20^{\circ}\), \(\cos 20^{\circ},\), and \(\tan 20^{\circ}\).

Look for \(20^{\circ}\) in the angle column of the table on page 356:

If you are using a pocket calculator, first make sure that it is in degree mode. Then type in 20 , followed by the sin, cos, or tan keys.

Answer: \(\sin 20^{\circ}=0.3420\), \(\cos 20^{\circ}=0.9397\), \(\tan 20^{\circ}=0.3640\)

Example \(\PageIndex{2}\)

Find \(x\) to the nearest tenth:

We wish to find the leg opposite \(20^{\circ}\) and we know the hypotenuse. We use the sine because it is the only one of the three trigonometric functions which involves both the opposite leg and the hypotenuse.

\(\begin{array} {rcl} {\sin 20^{\circ}} & = & {\dfrac{\text{opp}}{\text{hyp}}} \\ {.3420} & = & {\dfrac{x}{10}} \\ {(10)(.3420)} & = & {\dfrac{x}{\cancel{10}} \cancel{(10)}} \\ {3.420} & = & {x} \end{array}\)

If you are using a pocket calculator, type

Answer: \(x = 3.4\)

Example \(\PageIndex{3}\)

We know the hypotenuse and we wish to find the leg adjacent to \(\angle A\). We therefore use the cosine.

\(\begin{array} {rcl} {\cos 20^{\circ}} & = & {\dfrac{\text{adj}}{\text{hyp}}} \\ {0.9397} & = & {\dfrac{x}{10}} \\ {9.397} & = & {x} \end{array}\)

Answer: \(x = 9.4\)

We know the leg opposite \(\angle A\) and we wish to find the leg adjacent to \(\angle A\). We therefore use the tangent.

\(\begin{array} {rcl} {\tan 20^{\circ}} & = & {\dfrac{\text{opp}}{\text{adj}}} \\ {.3640} & = & {\dfrac{10}{x}} \\ {(x)(0.3640)} & = & {(\dfrac{10}{\cancel{x}})\cancel{(x)}} \\ {0.3640x} & = & {10} \\ {\dfrac{0.3640 x}{0.3640}} & = & {\dfrac{10}{0.3640}} \\ {x} & = & {\dfrac{10}{0.3640} = \dfrac{10}{0.364} = 27.47} \end{array}\)

27.47 is obtained by long division:

Answer: \(x = 27.5\)

There is an easier method to solve Example \(\PageIndex{4}\). \(\angle B = 90^{\circ}-20^{\circ} = 70^{\circ}\) The leg opposite \(\angle B\) is \(x\) and the leg adjacent to \(\angle B\) is 10 .

\(\begin{array} {rcl} {\tan 70^{\circ}} & = & {\dfrac{\text{opp}}{\text{adj}}} \\ {2.7475} & = & {\dfrac{x}{10}} \\ {(2.7475)(10)} & = & {x} \\ {27.475} & = & {x} \\ {27.5} & = & {x} \end{array}\)

This method is easier because it involves multiplication rather than long division.

Example \(\PageIndex{5}\)

\(\begin{array} {rcl} {\sin 14^{\circ}} & = & {\dfrac{\text{opp}}{\text{hyp}}} \\ {.2419} & = & {\dfrac{7}{x}} \\ {.2419 x} & = & {7} \\ {x} & = & {\dfrac{7}{.2419} = 28.9} \end{array}\)

In this case there is no way of avoiding long division.(It is possible to avoid long division by introducing tables for the secant and cosecant functions. We will not do so in this book.)

Answer: \(x = 28.9\)

Find \(x\) to the nearest degree:

Example \(\PageIndex{6}\)

\(\begin{array} {rcl} {\sin x^{\circ}} & = & {\dfrac{\text{opp}}{\text{hyp}}} \\ {\sin x^{\circ}} & = & {\dfrac{2}{3} = .6667} \end{array}\)

In the table we look in the sine column for the value closest to .6667:

.6667 is closest to .6691 because .6691 - .6667 = .0024 whereas .6667 - .6561 = .0106. Therefore \(x^{\circ} = 42^{\circ}\), to the nearest degree.

If you are using a pocket calculator, you will need to use the INV sin or 2nd F sin or SHIFT sin or \(sin^{-1}\) keys, depending on the model of calculatore. Type 2 \(\div\) 3 = INV sin, then round to the nearest degree.

Answer: \(x = 42\)

Example \(\PageIndex{7}\)

\(\begin{array} {rcl} {\sin 40^{\circ}} & = & {\dfrac{\text{opp}}{\text{hyp}}} \\ {.6428} & = & {\dfrac{x}{4}} \\ {(4)(.6428)} & = & {\dfrac{x}{\cancel{4}} (\cancel{4})} \\ {2.5712} & = & {x} \\ {2.6} & = & {x} \end{array}\)

Answer: \(x = 2.6\).

Example \(\PageIndex{8}\)

Find \(x\) and \(y\) to the nearest tenth:

\(\begin{array} {rcl} {\sin 65^{\circ}} & = & {\dfrac{\text{opp}}{\text{hyp}}} \\ {.9063} & = & {\dfrac{x}{8}} \\ {(8)(.9063)} & = & {\dfrac{x}{\cancel{8}} (\cancel{8})} \\ {7.2504} & = & {x} \\ {7.3} & = & {x} \end{array}\)

To find \(y\) we first find \(AD\):

\(\begin{array} {rcl} {\cos 65^{\circ}} & = & {\dfrac{\text{adj}}{\text{hyp}}} \\ {.4226} & = & {\dfrac{AD}{8}} \\ {(8)(.4226)} & = & {\dfrac{AD}{8} (8)} \\ {3.3808} & = & {AD} \end{array}\)

Since \(AC = BC = 8\) we have \(\angle A = \angle B = 65^{\circ}\). Therefore \(BD = AD = 3.3808\). \(y = AD + BD = 3.3808 + 3.3808 = 6.7616 = 6.8\).

Answer: \(x = 7.3, y = 6.8\).

Historical Note

The first table of trigonometric values was constructed by the Greek astronomer Hipparchus (c. 180-125 B.C.). Hipparchus assumed the vertex of each angle to be the center of a circle, as \(\angle AOB\) is shown to be in the circle of Figure \(\PageIndex{1}\). Depending on the number of degrees in \(\angle AOB\), his table would give the length of the chord \(AB\) relative to the radius of the circle. Today we would measure \(\angle AOC\) instead of \(\angle AOB\) and use the half chord \(AC\) instead of \(AB\). The ratio \(\dfrac{AC}{AO}\) is then just the sine of \(\angle AOC\).

Hipparchus obtained some of the values for his table from the properties of special geometric figures, such as the \(30^{\circ}-60^{\circ}-90^{\circ}\) triangle and the \(45^{\circ}-45^{\circ}-90^{\circ}\) triangle. The rest of the values were obtained from those already known by using trigonometric identities and approximation. The identities he used were essentially the half-angle and sum and difference formulas which students encounter in modern trigonometrJ courses.

The trigonometry of the Greeks, and later of the Hindus and the Arabs, was based primarily on the sine function. The Hindus replaced the table of chords of Hipparchus with a table of half chords. The term sine is derived from a Hindu word meaning "half-chord."

Gradually the right triangle replaced the chords of circles as the basis of trigonometric definitions. The cosine is just the sine of the complement of the angle in a right triangle, For example the complement of \(60^{\circ}\) is \(30^{\circ}\) and \(\cos 60^{\circ} = \sin 30^{\circ} = .5\).

A tangent is a line which touches a circle at only one point (see Chapter 7). In trigonometry it refers to just that part of the tangent line intercepted by the angle, relative to the radius of the circle. In Figure \(\PageIndex{2}\) the tangent of \(\angle DOE\) is the segment \(DE\) divided by the radius \(OD\). The ancient Greeks were probably aware of the tangent function but the first known table of values was constructed by the Arabs in 10th century. The term "tangent" was adopted in the 16th century.

Modern trigonometric tables are constructed from infinite series. These were first discovered in the 17th century by Newton, Leibniz and others. For example the infinite series for the sine function is

\[\sin x = x - \dfrac{x^3}{6} + \dfrac{x^5}{120} - \dfrac{x^7}{5040} + \cdots\]

where \(x\) is in radians, 1 radian = 57.296 degrees. A good approximation of the sine of an angle can be obtained from the infinite series by summing just the first few tenns. This is also the method computers and pocket calculators use to find trigonometric values. The derivation of these formulas is found in calculus textbooks.

1 - 10. Find each of the following using the table:

1. \(\sin 10^{\circ}\)

2. \(\sin 30^{\circ}\)

3. \(\cos 80^{\circ}\)

4. \(\cos 60^{\circ}\)

5. \(\tan 45^{\circ}\)

6. \(\tan 60^{\circ}\)

7. \(\sin 18^{\circ}\)

8. \(\cos 72^{\circ}\)

9. \(\tan 50^{\circ}\)

10. \(\tan 80^{\circ}\)

11 - 30. Find \(x\) to the nearest tenth:

31 - 38. Find \(x\) to the nearest degree:

39 - 46. Find \(x\) or \(x\) and \(y\) to the nearest tenth: