3.1 Obtuse Angles

The town of Avery lies 48 miles due east of Baker, and Clio is 34 miles from Baker, in the direction $35 degrees$ west of north. How far is it from Avery to Clio?

We know how to solve right triangles using the trigonometric ratios. But the triangle formed by the three towns is not a right triangle, because it includes an obtuse angle of $125 degrees$ at $B$ , as shown in the figure.

A triangle that is not a right triangle is called an oblique triangle. In this chapter we learn how to solve oblique triangles using the laws of sines and cosines. But first we must be able to find the sine, cosine, and tangent ratios for obtuse angles.

Angles in Standard Position

To extend our definition of the trigonometric ratios to obtuse angles, we use a Cartesian coordinate system. We put an angle $θ$ in standard position as follows:

Place the vertex at the origin with the initial side on the positive $x$ -axis;
the terminal side opens in the counter-clockwise direction.
We choose a point $P$ on the terminal side of the angle, and form a right triangle by drawing a vertical line from $P$ to the $x$ -axis.

The length of the side adjacent to $θ$ is the $x$ -coordinate of point $P$ , and the length of the side opposite is the $y$ -coordinate of $P$ . The length of the hypotenuse is the distance from the origin to $P$ , which we call $r$ . With this notation, our definitions of the trigonometric ratios are as follows.

It doesn't matter which point $P$ on the terminal side we use to calculate the trig ratios. If we choose some other point, say $P^{'}$ , with coordinates $(x^{'}, y^{'})$ , as shown at right, we will get the same values for the sine, cosine and tangent of $θ$ . The new triangle formed is similar to the first one, so the ratios of the sides of the new triangle are equal to the corresponding ratios in the first triangle.

Example 1

Find the values of $\cos (θ), \sin (θ)$ , and $\tan (θ)$ if the point $open parenthesis 12, 5 close parenthesis$ is on the terminal side of $θ$ .

For the point $P open parenthesis 12, 5 close parenthesis$ , we have $x equals 12$ and $y equals 5$ . We use the distance formula to find $r$ .

$\begin{aligned} r & = \sqrt{(2 - 0)^{2} + (5 - 0)^{2}} \\ = \sqrt{25 + 144} = \sqrt{169} = 13 \end{aligned}$

The trig ratios are

$\begin{aligned} \cos (θ) & = \frac{x}{r} = \frac{12}{13} \\ \sin (θ) & = \frac{y}{r} = \frac{5}{13} \\ \tan (θ) & = \frac{y}{x} = \frac{5}{12} \end{aligned}$

Find the equation of the terminal side of the angle in the previous example. (Hint: The terminal side lies on a line that goes through the origin and the point $open parenthesis 12, 5 close parenthesis$ .)
Show that the point $P prime open parenthesis 24, 10 close parenthesis$ also lies on the terminal side of the angle.
Compute the trig ratios for $θ$ using the point $P^{'}$ instead of $P$ .

$y = \frac{5}{12} x$
$open parenthesis 24, 10 close parenthesis$ satisfies $y = \frac{5}{12} x$ , that is, the equation $10 = \frac{5}{12} (24)$ is true.
$r^{2} = 24^{2} + 10^{2} = 676$ , so $r = \sqrt{676} = 26.$ Then
$\begin{matrix} \cos (θ) = \frac{x}{r} = \frac{24}{26} = \frac{12}{13} \\ \sin (θ) = \frac{y}{r} = \frac{10}{26} = \frac{5}{13} \\ \tan (θ) = \frac{y}{x} = \frac{10}{24} = \frac{5}{12} \end{matrix}$

Activity

Obtuse Angles

Recall that an obtuse angle has measure between $90 degrees$ and $180 degrees$ . Use the coordinate definitions of the trig ratios described above to complete the Activity.

1. Draw an angle $θ$ in standard position with the point $P open parenthesis 6, 4 close parenthesis$ on its terminal side.
2. Find $r$ , the distance from the origin to $P$ .
3. Calculate $\sin (θ), \cos (θ)$ , and $\tan (θ)$ . Give both exact answers and decimal approximations rounded to four places.
4. Use the inverse cosine key on your calculator to find $θ$ . Use your calculator to verify the values of $\sin (θ), \cos (θ)$ , and $\tan (θ)$ that you found in part (3).
1. Draw another angle $ϕ$ in standard position with the point $Q open parenthesis minus 6, 4 close parenthesis$ on its terminal side.
2. Explain why $ϕ$ is the supplement of $θ$ . (Hint: Consider the right triangles formed by drawing vertical lines from $P$ and $Q$ .)
3. Can you use the right triangle definitions (using opposite, adjacent and hypotenuse) to compute the sine and cosine of $ϕ$ ? Why or why not?
4. Calculate $\sin (ϕ), \cos (ϕ)$ , and $\tan (ϕ)$ using the coordinate definitions. How are the trig values of $ϕ$ related to the trig values of $θ$ ?
1. Explain why $θ$ and $ϕ$ have the same sine but different cosines.
2. Use the inverse cosine key on your calculator to find $ϕ$ . Use your calculator to verify the values of $\sin (ϕ), \cos (ϕ)$ , and $\tan (ϕ)$ that you found above.
3. Compute $180 degrees minus ϕ$ . What answer should you expect to get?

Trigonometric Ratios for Obtuse Angles

Our new definitions for the trig ratios work just as well for obtuse angles, even though $θ$ is not technically “inside” a triangle, because we use the coordinates of $P$ instead of the sides of a triangle to compute the ratios.

Notice first of all that because $x$ -coordinates are negative in the second quadrant, the cosine and tangent ratios are both negative for obtuse angles. For example, in the figure below, the point $open parenthesis minus 4, 3 close parenthesis$ lies on the terminal side of the angle $θ$ . We see that $r = \sqrt{(- 4)^{2} + 3^{2}} = 5$ , so

$\begin{aligned} \cos (θ) & = \frac{x}{r} = \frac{- 4}{5} \\ \sin (θ) & = \frac{y}{r} = \frac{3}{5} \\ \tan (θ) & = \frac{y}{x} = \frac{3}{- 4} = \frac{- 3}{4} \end{aligned}$

Example 2

Find the values of $\cos (θ)$ and $\tan (θ)$ if $θ$ is an obtuse angle with $\sin (θ) = \frac{1}{3}$ .

Because $θ$ is obtuse, the terminal side of the angle lies in the second quadrant, as shown in the figure below. Because $\sin (θ) = \frac{1}{3}$ , we know that $\frac{y}{r} = \frac{1}{3}$ , so we can choose a point $P$ with $y equals 1$ and $r equals 3$ . To find $\cos (θ)$ and $\tan (θ)$ we need to know the value of $x$ . From the Pythagorean Theorem,

$\begin{aligned} x^{2} + 1^{2} & = 3^{2} \\ x^{2} & = 3^{2} - 1^{2} = 8 \\ x & = - \sqrt{8} \end{aligned}$

Remember that $x$ is negative in the second quadrant! Thus

$\cos (θ) = \frac{x}{r} = \frac{- \sqrt{8}}{3} and \tan (θ) = \frac{y}{x} = \frac{- 1}{\sqrt{8}}$

Sketch an obtuse angle $θ$ whose cosine is $\frac{- 8}{17}$ .
Find the sine and the tangent of $θ$ .

By the Pythagorean theorem,
$y^{2} = 17^{2} - (- 8)^{2} = 15^{2}$
, so $y equals 15$ .
$\sin (θ) = \frac{15}{17}, \tan (θ) = \frac{- 15}{8}$

Using a Calculator

In the Examples above, we used a point on the terminal side to find exact values for the trigonometric ratios of obtuse angles. Scientific and graphing calculators are programmed with approximations for these trig ratios.

Example 3

Find the sine and cosine of $130 degrees$ . Compare to the sine and cosine of $50 degrees$ .

Using a calculator and rounding the values to four places, we find

$\begin{aligned} \sin (130^{\circ}) & = 0.7660 and \cos (130^{\circ}) = - 0.6428 \\ \sin (50^{\circ}) & = 0.7660 and \cos (50^{\circ}) = 0.6428 \end{aligned}$

We see that $sin ⁡ open parenthesis 130 degrees close parenthesis equals sin ⁡ open parenthesis 50 degrees close parenthesis$ and $\cos (130^{\circ}) = - \cos (50^{\circ})$ . This result should not be surprising when we look at both angles in standard position, as shown below.

The angles $50 degrees$ and $130 degrees$ are supplementary. The right triangles formed by choosing the points $(x, y)$ and $(- x, y)$ on their terminal sides are congruent triangles.

Consequently, the trigonometric ratios for $50 degrees$ and for $130 degrees$ are equal, except that the cosine of $130 degrees$ is negative.

Use your calculator to fill in the table. Round to four decimal places.

$θ$	$\cos (θ)$	$\sin (θ)$	$180 degrees minus θ$	$cos ⁡ open parenthesis 180 degrees minus θ close parenthesis$	$sin ⁡ open parenthesis 180 degrees minus θ close parenthesis$
$10 degrees$
$20 degrees$
$30 degrees$
$40 degrees$
$50 degrees$
$60 degrees$
$70 degrees$
$80 degrees$

$θ$	$\cos (θ)$	$\sin (θ)$	$180 degrees minus θ$	$cos ⁡ open parenthesis 180 degrees minus θ close parenthesis$	$sin ⁡ open parenthesis 180 degrees minus θ close parenthesis$
$10 degrees$	$0.9848$	$0.1736$	$170 degrees$	$minus 0.9848$	$0.1736$
$20 degrees$	$0.9397$	$0.3420$	$160 degrees$	$minus 0.9397$	$0.3420$
$30 degrees$	$0.8660$	$0.5$	$150 degrees$	$0.8660$	$minus 0.5$
$40 degrees$	$0.7660$	$0.6428$	$140 degrees$	$minus 0.7660$	$0.6428$
$50 degrees$	$0.6428$	$0.7660$	$130 degrees$	$minus 0.6428$	$0.7660$
$60 degrees$	$0.5$	$0.8660$	$120 degrees$	$minus 0.5$	$0.8660$
$70 degrees$	$0.3420$	$0.9397$	$110 degrees$	$minus 0.9397$	$0.3420$
$80 degrees$	$0.1736$	$0.9848$	$100 degrees$	$minus 0.9848$	$0.1736$

Trigonometric Ratios for Supplementary Angles

The Examples above illustrate the following equations for supplementary angles. These three equations are called identities, which means that they are true for all values of the variable $θ$ .

Example 4

Find two different angles $θ$ , rounded to the nearest $0.1 degrees$ , that satisfy $sin ⁡ open parenthesis θ close parenthesis equals 0.25$ .

To find an angle with $sin ⁡ open parenthesis θ close parenthesis equals 0.25$ , we calculate $θ = \sin^{- 1} (0.25)$ . With the calculator in degree mode, we press

2nd SIN 0.25 ENTER

and find that one angle is $θ approximately equals 14.5 degrees$ . We draw this acute angle in standard position in the first quadrant, and sketch in a right triangle as shown below. There must also be an obtuse angle whose sine is $0.25$ . To see the second angle, we draw a congruent triangle in the second quadrant as shown.

The supplement of $14.5 degrees$ , namely $θ = 180^{\circ} - {14.5}^{\circ} = {165.5}^{\circ}$ , is the obtuse angle we need. Because the ratio $\frac{y}{r}$ is the same for both triangles, they have the same sine.

Find two different angles $θ$ that satisfy $sin ⁡ open parenthesis θ close parenthesis equals 0.8$ .

One angle is $θ = \sin^{- 1} (0.8) = {53.13}^{\circ}$ . The second angle is the supplement of $53.13 degrees$ , or $(180 - 53.13) = {126.87}^{\circ}$ .

Because there are two angles with the same sine, it is easier to find an obtuse angle if we know its cosine instead of its sine.

Caution

In the previous example, you might notice that $\tan (θ) = \frac{- 4}{3}$ and try to find by calculating $\tan^{- 1} (\frac{- 4}{3})$ . However, if we press

2nd TAN 4 3 ENTER

the calculator returns an angle of $θ \approx - {53.1}^{\circ}$ . It is true that $\tan (- {53.1}^{\circ}) = \frac{- 4}{3}$ , but this is not the obtuse angle we want.

We also know that $\sin θ = \frac{4}{5}$ , and if we press

2nd SIN 4 5 ENTER

we get $θ approximately equals 53.1 degrees$ .

This is the acute angle whose terminal side passes through the point $open parenthesis 3, 4 close parenthesis$ , as shown in the figure above. The angle we want is its supplement, $θ \approx 180^{\circ} - {53.1}^{\circ} = {126.9}^{\circ}$ .

Find the cosine of an obtuse angle with $\tan (θ) = - 2$ .
Find the angle $θ$ in part (a).

The point $open parenthesis minus 1, 2 close parenthesis$ lies on the terminal side of the angle, so
$r^{2} = (- 1)^{2} + 2^{2} = 5$
and $r = \sqrt{5}$ . Then $\cos (θ) = \frac{- 1}{\sqrt{5}}$ .
$θ = \cos^{- 1} (\frac{- 1}{\sqrt{5}}) \approx {116.565}^{\circ}$

Supplements of the Special Angles

In Chapter 2 we learned that the angles $30 degrees, 45 degrees$ and $60 degrees$ are useful because we can find exact values for their trigonometric ratios. The same is true for the supplements of these angles in the second quadrant, shown at right.

Example 6

Find exact values for the trigonometric ratios of $135 degrees$ .

We sketch an angle of $θ equals 135 degrees$ in standard position, as shown below. The terminal side is in the second quadrant and makes an acute angle of $45 degrees$ with the negative $x$ -axis, and passes through the point $open parenthesis minus 1, 1 close parenthesis$ . Thus, $r = \sqrt{(- 1)^{2} + 1^{2}} = \sqrt{2}$ , and we calculate

$\begin{aligned} \cos (135^{\circ}) & = \frac{x}{r} = \frac{- 1}{\sqrt{2}} \\ \sin (135^{\circ}) & = \frac{y}{r} = \frac{1}{\sqrt{2}} \\ \tan (135^{\circ}) & = \frac{y}{x} = \frac{1}{- 1} = - 1 \end{aligned}$

Find exact values for the trigonometric ratios of $120 degrees$ and $150 degrees$ .

$120 degrees$ is the supplement of $60 degrees$ , and $150 degrees$ is the supplement of $30 degrees$ .

$θ$	$\cos (θ)$	$\sin (θ)$	$\tan (θ)$
$120 degrees$	$\frac{- 1}{2}$	$\frac{\sqrt{3}}{2}$	$- \sqrt{3}$
$150 degrees$	$\frac{- \sqrt{3}}{2}$	$\frac{1}{2}$	$\frac{- 1}{\sqrt{3}}$

We can also find the trig ratios for the quadrantal angles. These are the angles, including $0 degrees$ , $90 degrees$ and $180 degrees$ , whose terminal sides lie on one of the axes.

Example 7

Find exact values for the trigonometric ratios of $90 degrees$ .

The terminal side of a $90 degrees$ angle in standard position is the positive $y$ -axis. If we take the point $P open parenthesis 0, 1 close parenthesis$ on the terminal side as shown at right, then $x equals 0$ and $y equals 1$ . Although we don't have a triangle, we can still calculate a value for $r$ , the distance from the origin to $P$ .

$r = \sqrt{0^{2} + 1^{2}} = 1$

Our coordinate definitions for the trig ratios give us

$\cos (90^{\circ}) = \frac{x}{r} = \frac{0}{1} and \sin (90^{\circ}) = \frac{y}{r} = \frac{1}{1}$

so $cos ⁡ open parenthesis 90 degrees close parenthesis equals 0$ and $sin ⁡ open parenthesis 90 degrees close parenthesis equals 1.$ Also, $\tan (90^{\circ}) = \frac{y}{x} = \frac{1}{0},$ so $tan ⁡ open parenthesis 90 degrees close parenthesis$ is undefined.

Find exact values for the trigonometric ratios of $180 degrees$ .

The point $open parenthesis minus 1, 0 close parenthesis$ lies on the terminal side of the angle $180 degrees$ . Thus

$\cos (180^{\circ}) = - 1, \sin (180^{\circ}) = 0, \tan (180^{\circ}) = 0$

The Area of a Triangle

The figure below shows part of the map for a new housing development, Pacific Shores. You are interested in the corner lot, number 86, and you would like to know the area of the lot in square feet. The sales representative for Pacific Shores provides you with the dimensions of the lot, but you don't know a formula for the area of an irregularly shaped quadrilateral.

It occurs to you that you can divide the quadrilateral into two triangles, and find the area of each. Now, you know a formula for the area of a triangle in terms of its base and height, namely,

$A = \frac{1}{2} b h$

but unfortunately, you don't know the height of either triangle.

However, you can easily measure the angles at the corners of the lot using the plot map and a protractor. You can check the values on the plot map for lot 86 shown above.

Using trigonometry, we can find the area of a triangle if we know two of its sides, say $a$ and $b$ , and the included angle, $θ$ . The figure below shows three possibilities, depending on whether the angle $θ$ is acute, obtuse, or $90 degrees$ .

In each case, $b$ is the base of the triangle, and its altitude is $h$ . Our task is to find an expression for $h$ in terms of the quantities we know: $a$ , $b$ , and $θ$ . You should check that in all three triangles

$\sin (θ) = \frac{h}{a}$

Solving for $h$ gives us $h = a \sin (θ)$ . Finally, we substitute this expression for $h$ into our old formula for the area to get

$A = \frac{1}{2} b h = \frac{1}{2} b a \sin (θ)$

Example 8

Find the area of lot 86.

For the triangle in the lower portion of lot 86, $a equals 120.3$ , $b equals 141$ , and $θ equals 95 degrees$ . The area of that portion is

$\begin{aligned} First Area & = \frac{1}{2} a b \sin (θ) \\ = \frac{1}{2} (120.3) ((141) \sin (95^{\circ}) \approx 8448.88 \end{aligned}$

For the triangle in the upper portion of the lot, $a equals 161$ , $b equals 114.8$ , and $θ equals 86.1 degrees$ . The area of that portion is

$\begin{aligned} Second Area & = \frac{1}{2} a b \sin (θ) \\ = \frac{1}{2} (161) ((114.8) \sin ({86.1}^{\circ}) \approx 9220.00 \end{aligned}$

The total area of the lot is the sum of the areas of the triangles

$Total area = First Area + Second Area \approx 17668.88$

Lot 86 has an area of approximately 17,669 square feet.

A triangle has sides of length 6 and 7, and the angle between those sides is $150 degrees$ . Find the area of the triangle.

The area is given by

$A = \frac{1}{2} (6) (7) \sin (150^{\circ}) = 21 (\frac{1}{2}) = \frac{21}{2}$

Review the following skills you will need for this section.

Section 3.1 Summary

Vocabulary

Standard position
Initial side
Terminal side
Quadrantal angle
Oblique triangle
Quadrilateral
Identity

Concepts

We put an angle in standard position by placing its vertex at the origin and the initial side on the positive $x$ -axis.
Note
Coordinate Definitions of the Trigonometric Ratios
- $\cos (θ) = \frac{x}{r}$
- $\sin (θ) = \frac{y}{r}$
- $\tan (θ) = \frac{y}{x}$
Note
Trigonometric Ratios for Supplementary Angles
- $\cos (180^{\circ} - θ) = - \cos θ$
- $\sin (180^{\circ} - θ) = \sin θ$
- $\tan (180^{\circ} - θ) = - \tan θ$
There are always two (supplementary) angles between $0 degrees$ and $180 degrees$ that have the same sine. Your calculator will only tell you one of them.
Note
Area of a Triangle
If a triangle has sides of length $a$ and $b$ , and the angle between those two sides is $θ$ , then the area of the triangle is given by
$A = \frac{1}{2} a b \sin (θ)$

Study Questions

Delbert says that $\sin (θ) = \frac{4}{7}$ in the figure. Is he correct? Why or why not?
Give the lengths of the legs of each right triangle.
a.
b.
Explain why the length of the horizontal leg of the right triangle is $- x$ .
Why are the sines of supplementary angles equal, but the cosines are not? What about the tangents of supplementary angles?
Use your calculator to evaluate $sin ⁡ open parenthesis 118 degrees close parenthesis$ , then evaluate $\sin^{- 1} (ANS)$ . Explain the result.
Write an expression for the area of the triangle.

Skills

Practice each skill in the Homework Problems listed.

Use the coordinate definition of the trig ratios #3-20, 45-48
Find the trig ratios of supplementary angles #7-10, 21-38
Know the trig ratios of the special angles in the second quadrant #21, 41-44
Find two solutions of the equation $\sin (θ) = k$ #29-38
Find the area of a triangle #49-58

Homework 3.1

Without using pencil and paper or a calculator, give the supplement of each angle.

$30 degrees$
$45 degrees$
$120 degrees$
$25 degrees$
$165 degrees$
$110 degrees$

$150 degrees$
$135 degrees$
$60 degrees$
$155 degrees$
$15 degrees$
$70 degrees$

Without using pencil and paper or a calculator, give the complement of each angle.

$60 degrees$
$80 degrees$
$25 degrees$
$18 degrees$
$64 degrees$
$47 degrees$

For Problems 3–6,

Give the coordinates of point $P$ on the terminal side of the angle.
Find the distance from the origin to point $P$ .
Find $\cos (θ), \sin (θ),$ and $\tan (θ) .$

$open parenthesis 5, 2 close parenthesis$
$the square root of 29$
$\cos (θ) = \frac{5}{\sqrt{29}}, \sin (θ) = \frac{2}{\sqrt{29}}, \tan (θ) = \frac{2}{5}$

$open parenthesis minus 4, 7 close parenthesis$
$the square root of 65$
$\cos (θ) = \frac{- 4}{\sqrt{65}}, \sin (θ) = \frac{7}{\sqrt{65}}, \tan (θ) = \frac{- 7}{4}$

For Problems 7–10,

Find the sine and cosine of the angle.
Sketch the supplement of the angle in standard position. (Use congruent triangles.)
Find the sine and cosine of the supplement.
Find the angle and its supplement, rounded to the nearest degree.

$\sin (θ) = \frac{9}{\sqrt{97}}$ , $\cos (θ) = \frac{4}{\sqrt{97}}$
$\sin (180^{\circ} - θ) = \frac{9}{\sqrt{97}}$ , $\cos (180^{\circ} - θ) = \frac{- 4}{\sqrt{97}}$
$θ = 66^{\circ}, 180^{\circ} - θ = 114^{\circ}$

$\sin (θ) = \frac{8}{\sqrt{89}}$ , $\cos (θ) = \frac{- 5}{\sqrt{89}}$
$\sin (180^{\circ} - θ) = \frac{8}{\sqrt{89}}$ , $\cos (180^{\circ} - θ) = \frac{5}{\sqrt{89}}$
$θ = 122^{\circ}, 180^{\circ} - θ = 58^{\circ}$

For Problems 11–20,

Sketch an angle in standard position with the given properties.
Find $\cos (θ), \sin (θ),$ and $\tan (θ) .$
Find the angle $θ$ , rounded to tenths of a degree.

The point $open parenthesis minus 5, 12 close parenthesis$ is on the terminal side.

$\cos (θ) = \frac{- 5}{13}$ , $\sin (θ) = \frac{12}{13}$ , $\tan (θ) = \frac{- 12}{5}$
$112.6 degrees$

The point $open parenthesis 12, 9 close parenthesis$ is on the terminal side.

$\cos θ = - 0.8$

$\cos (θ) = \frac{3}{5}, \tan (θ) = \frac{- 3}{4}$
$143.1 degrees$

$\cos (θ) = \frac{5}{13}$

$\cos (θ) = \frac{3}{11}$

$\sin (θ) = \frac{\sqrt{112}}{11}$ , $\tan (θ) = \frac{\sqrt{112}}{3}$
$74.2 degrees$

$\cos (θ) = \frac{- 5}{6}$

$\tan (θ) = \frac{- 1}{6}$

$\sin (θ) = \frac{1}{\sqrt{37}}$ , $\cos (θ) = \frac{- 6}{\sqrt{37}}$
$170.5 degrees$

$\tan (θ) = \frac{9}{5}$

$tan ⁡ open parenthesis θ close parenthesis equals 4$

$\sin (θ) = \frac{4}{\sqrt{17}}$ , $\cos (θ) = \frac{1}{\sqrt{17}}$
$76.0 degrees$

$\tan (θ) = - 1$

Fill in exact values from memory without using a calculator.

$θ$	$0 degrees$	$30 degrees$	$45 degrees$	$60 degrees$	$90 degrees$	$120 degrees$	$135 degrees$	$150 degrees$	$180 degrees$
$\cos (θ)$
$\sin (θ)$
$\tan (θ)$

$θ$	$0 degrees$	$30 degrees$	$45 degrees$	$60 degrees$	$90 degrees$	$120 degrees$	$135 degrees$	$150 degrees$	$180 degrees$
$\cos (θ)$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	$0$	$\frac{- 1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{- \sqrt{3}}{2}$	$minus 1$
$\sin (θ)$	$0$	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	$0$
$\tan (θ)$	$0$	$\frac{1}{\sqrt{3}}$	$1$	$the square root of 3$	$undefined$	$- \sqrt{3}$	$minus 1$	$\frac{- 1}{\sqrt{3}}$	$0$

Use your calculator to fill in the table. Round values to four decimal places.

$θ$	$15 degrees$	$25 degrees$	$65 degrees$	$75 degrees$	$105 degrees$	$115 degrees$	$155 degrees$	$165 degrees$
$\cos (θ)$
$\sin (θ)$
$\tan (θ)$

For each angle $θ$ in the table for Problem 22, the angle $180 degrees minus θ$ is also in the table.

What is true about $\sin (θ)$ and $sin ⁡ open parenthesis 180 degrees minus θ close parenthesis$ ?
What is true about $\cos (θ)$ and $cos ⁡ open parenthesis 180 degrees minus θ close parenthesis$ ?
What is true about $\tan (θ)$ and $tan ⁡ open parenthesis 180 degrees minus θ close parenthesis$ ?

$\sin (θ) = \sin (180^{\circ} - θ)$
$\cos (θ) = - \cos (180^{\circ} - θ)$
$\tan (θ) = - \tan (180^{\circ} - θ)$

Describe and explain any patterns of equal values you see in the table for Problem 22.

For Problems 25–28,

Evaluate each pair of angles to the nearest $0.1 degrees$ , and show that they are supplements.
Sketch both angles.
Find the sine of each angle.

$θ = \cos^{- 1} (\frac{3}{4})$ , $ϕ = \cos^{- 1} (\frac{- 3}{4})$

$θ \approx {41.4}^{\circ}, ϕ \approx {138.6}^{\circ}$
$\sin (θ) = \sin (ϕ) = \frac{\sqrt{7}}{4}$

$θ = \cos^{- 1} (\frac{1}{5})$ , $ϕ = \cos^{- 1} (\frac{- 1}{5})$

$θ = \cos^{- 1} (0.1525)$ , $ϕ = \cos^{- 1} (- 0.1525)$

$θ \approx {81.2}^{\circ}, ϕ \approx {98.8}^{\circ}$
$\sin (θ) = \sin (ϕ) = \frac{\sqrt{156279}}{400} \approx 0.9883$

$θ = \cos^{- 1} (0.6825)$ , $ϕ = \cos^{- 1} (- 0.6825)$

For Problems 29–34, find two different angles that satisfy the equation. Round to the nearest $0.1 degrees$ .

$sin ⁡ open parenthesis θ close parenthesis equals 0.7$

$44.4 degrees$ and $135.6 degrees$

$sin ⁡ open parenthesis θ close parenthesis equals 0.1$

$\frac{\sin (θ)}{6} = 0.14$

$57.1 degrees$ and $122.9 degrees$

$\frac{5}{\sin (θ)} = 6$

$4.8 = \frac{3.2}{\sin (θ)}$

$41.8 degrees$ and $138.2 degrees$

$1.5 = \frac{\sin (θ)}{0.3}$

For Problems 35–38, fill in the blanks with complements or supplements.

If $sin ⁡ open parenthesis 57 degrees close parenthesis equals q$ , then $\sin (\underset{―}{}) = q$ also, $\cos (\underset{―}{}) = q$ , and $\cos (\underset{―}{}) = - q$ .

$sin ⁡ open parenthesis 123 degrees close parenthesis equals q$ , $cos ⁡ open parenthesis 33 degrees close parenthesis equals q$ , $\cos (147^{\circ}) = - q$

If $sin ⁡ open parenthesis 18 degrees close parenthesis equals w$ , then $\sin (\underset{―}{}) = w$ also, $\cos (\underset{―}{}) = w$ , and $\cos (\underset{―}{}) = - w$ .

If $cos ⁡ open parenthesis 74 degrees close parenthesis equals m$ , then $\cos (\underset{―}{}) = - m$ , and $\sin (\underset{―}{})$ and $\sin (\underset{―}{})$ both equal $m$ .

$\cos (106^{\circ}) = - m$ , $sin ⁡ open parenthesis 16 degrees close parenthesis equals m$ , $sin ⁡ open parenthesis 164 degrees close parenthesis equals m$

If $cos ⁡ open parenthesis 36 degrees close parenthesis equals t$ , then $\cos (\underset{―}{}) = - t$ , and $\sin (\underset{―}{})$ and $\sin (\underset{―}{})$ both equal $t$ .

Sketch the line $y = \frac{3}{4} x$ .
Find two points on the line with positive $x$ -coordinates.
The line $y = \frac{3}{4} x$ makes an angle with the positive $x$ -axis. What is that angle?
Repeat parts (a) through (c) for the line $y = \frac{- 3}{4} x$ , except find two points with negative $x$ -coordinates.

$open parenthesis 4, 3 close parenthesis, open parenthesis 8, 6 close parenthesis$
$y = \tan^{- 1} (\frac{3}{4}) \approx {36.87}^{\circ}$
$(- 4, 3), (- 8, 6); {143.13}^{\circ}$

Sketch the line $y = \frac{5}{3} x$ .
Find two points on the line with positive $x$ -coordinates.
The line $y = \frac{5}{3} x$ makes an angle with the positive $x$ -axis. What is that angle?
Repeat parts (a) through (c) for the line $y = \frac{- 5}{3} x$ , except find two points with negative $x$ -coordinates.

For Problems 41–44,

Find exact values for the base and height of the triangle.
Compute an exact value for the area of the triangle.

$b equals 8$ in, $h = 3 \sqrt{3}$ in
$12 the square root of 3$ sq in

$b = 6 - \frac{3 \sqrt{2}}{2}$ mi, $h = \frac{3 \sqrt{2}}{2}$ mi
$\frac{18 \sqrt{2} - 9}{4}$ sq mi

Sketch an angle of $120 degrees$ in standard position. Find the missing coordinates of the points on the terminal side.

$open parenthesis minus 1, ? close parenthesis$
$open parenthesis ?, 3 close parenthesis$

$(- 1, \sqrt{3})$
$(- \sqrt{3}, 3)$

Sketch an angle of $150 degrees$ in standard position. Find the missing coordinates of the points on the terminal side.

$open parenthesis ?, 2 close parenthesis$
$open parenthesis minus 4, ? close parenthesis$

Sketch an angle of $135 degrees$ in standard position. Find the missing coordinates of the points on the terminal side.

$open parenthesis ?, 3 close parenthesis$
$(- \sqrt{5}, ?)$

$open parenthesis minus 3, 3 close parenthesis$
$(- \sqrt{5}, \sqrt{5})$

Use a sketch to explain why $cos ⁡ open parenthesis 90 degrees close parenthesis equals 0$ .
Use a sketch to explain why $cos ⁡ open parenthesis 180 degrees close parenthesis equals 1$ .

For Problems 49–54, find the area of the triangle with the given properties. Round your answer to two decimal places.

$20.71$ sq m

$55.51$ sq cm

$a equals 0.8$ m, $c equals 0.15$ m, $B equals 15 degrees$

Find the area of the regular pentagon shown at right. (Hint: The pentagon can be divided into five congruent triangles.)

$38.04$ sq units

Find the area of the regular hexagon shown at right. (Hint: The hexagon can be divided into six congruent triangles.)

For Problems 57 and 58, lots from a housing development have been subdivided into triangles. Find the total area of each lot by computing and adding the areas of each triangle.

$13, 851.3$ sq ft

For Problems 59 and 60,

Find the coordinates of point $P$ . Round to two decimal places.
Find the sides $B C$ and $P C$ of $△ P C B$ .
Find side $P B$ .

$open parenthesis minus 74.97, 59.00 close parenthesis$
$B C = 141.97, P C = 59.00$
$153.74$

Later we will be able to show that $\sin (18^{\circ}) = \frac{\sqrt{5} - 1}{4}$ . What is the exact value of $sin ⁡ open parenthesis 162 degrees close parenthesis ?$ (Hint: Sketch both angles in standard position.)

$\frac{\sqrt{5} - 1}{4}$

Later we will be able to show that $\cos (36^{\circ}) = \frac{\sqrt{5} + 1}{4}$ . What is the exact value of $cos ⁡ open parenthesis 144 degrees close parenthesis ?$ (Hint: Sketch both angles in standard position.)

Alice wants an obtuse angle $θ$ that satisfies $sin ⁡ open parenthesis θ close parenthesis equals 0.3$ . Bob presses some buttons on his calculator and reports that $θ equals 17.46 degrees$ . Explain Bob's error and give a correct approximation of $θ$ accurate to two decimal places.

Bob found an acute angle. The obtuse angle is the supplement of $17.46 degrees$ , or $162.54 degrees$ .

Yaneli finds that the angle $θ$ opposite the longest side of a triangle satisfies $sin ⁡ open parenthesis θ close parenthesis equals 0.8$ . Zelda reports that $θ equals 53.13 degrees$ . Explain Zelda's error and give a correct approximation of $θ$ accurate to two decimal places.

For Problems 65–70,

Sketch an angle $θ$ in standard position, $0 degrees less than or equal to θ less than or equal to 180 degrees$ , with the given properties.
Find expressions for $\cos (θ), \sin (θ)$ , and $\tan (θ)$ in terms of the given variable.

$\cos (θ) = \frac{x}{3}, x < 0$

$\cos (θ) = \frac{x}{3}$ , $\sin (θ) = \frac{\sqrt{9 - x^{2}}}{3}$ , $\tan (θ) = \frac{\sqrt{9 - x^{2}}}{x}$

$\tan (θ) = \frac{4}{α}, α < 0$

$θ$ is obtuse and $\sin (θ) = \frac{y}{2}$

$\cos (θ) = \frac{- \sqrt{4 - y^{2}}}{2}$ , $\sin (θ) = \frac{y}{2}$ , $\tan (θ) = \frac{- y}{\sqrt{4 - y^{2}}}$

$θ$ is obtuse and $\tan (θ) = \frac{q}{- 7}$

$θ$ is obtuse and $\tan (θ) = m$

$\cos (θ) = \frac{- 1}{\sqrt{1 + m^{2}}}$ , $\sin (θ) = \frac{- m}{\sqrt{1 + m^{2}}}$ , $\tan (θ) = m$

$\cos (θ) = h$

Trigonometry by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.