How many degrees are in each fraction of one complete revolution?
5. [latex]\dfrac{1}{4}[/latex]
6. [latex]\dfrac{1}{5}[/latex]
7. [latex]\dfrac{1}{6}[/latex]
8. [latex]\dfrac{1}{8}[/latex]
Algebra Refresher Answers
[latex]24[/latex]
[latex]24[/latex]
[latex]24[/latex]
[latex]24[/latex]
[latex]90°[/latex]
[latex]72°[/latex]
[latex]60°[/latex]
[latex]45°[/latex]
Learning Objectives
Use the coordinate definition of the trig ratios.
Find the trig ratios of supplementary angles.
Know the trig ratios of the special angles in the second quadrant.
Find two solutions of the equation [latex]\sin \theta = k[/latex].
Find the area of a triangle.
The town of Avery lies [latex]48[/latex] miles due east of Baker, and Clio is [latex]34[/latex] miles from Baker, in the direction [latex]35^{\circ}[/latex] 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 [latex]125^{\circ}[/latex] at [latex]B[/latex], 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 [latex]\theta[/latex] in standard position as follows:
Place the vertex at the origin with the initial side on the positive [latex]x[/latex]-axis;
the terminal side opens in the counterclockwise direction.
We choose a point [latex]P[/latex] on the terminal side of the angle and form a right triangle by drawing a vertical line from [latex]P[/latex] to the [latex]x[/latex]-axis.
The length of the side adjacent to [latex]\theta[/latex] is the [latex]x[/latex]-coordinate of point [latex]P[/latex], and the length of the side opposite is the [latex]y[/latex]-coordinate of [latex]P[/latex]. The length of the hypotenuse is the distance from the origin to [latex]P[/latex], which we call [latex]r[/latex]. With this notation, our definitions of the trigonometric ratios are as follows.
Coordinate Definitions of the Trigonometric Ratios.
[latex]\cos \theta = \frac{x}{r}[/latex]
[latex]\sin \theta = \frac{y}{r}[/latex]
[latex]\tan \theta = \frac{y}{x}[/latex]
It doesn’t matter which point [latex]P[/latex] on the terminal side we use to calculate the trig ratios. If we choose some other point, say [latex]P'[/latex] with coordinates [latex]x', y'[/latex], as shown at right, we will get the same values for the [latex]\sin, \cos,[/latex] and tangent of [latex]\theta[/latex]. 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 3.1.
Find the values of [latex]\cos \theta[/latex], [latex]\sin \theta[/latex], and [latex]\tan \theta[/latex] if the point [latex](12,5)[/latex] is on the terminal side of [latex]\theta[/latex].
Solution
For the point [latex]P(12,5)[/latex], we have [latex]x= 12[/latex]and [latex]y=5[/latex]. We use the distance formula to find [latex]r[/latex]:
In the previous example, we get the same results by using the triangle definitions of the trig ratios. We create a right triangle by dropping a vertical line from [latex]P[/latex] to the [latex]x[/latex]-axis, as shown in the figure. The legs of the right triangle have lengths [latex]12[/latex] and [latex]5[/latex], and the hypotenuse has length [latex]13[/latex].
We could also use the Pythagorean Theorem to find the hypotenuse in the previous example.
Checkpoint 3.3.
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 [latex](12,5)[/latex].)
Show that the point [latex]P'(24, 10)[/latex] also lies on the terminal side of the angle.
Compute the trig ratios for [latex]\theta[/latex] using the point [latex]P'[/latex] instead of [latex]P[/latex]
Solution
[latex]y = \dfrac{5}{12}x[/latex].
[latex](24,10)[/latex] satisfies [latex]y = \dfrac{5}{12}x[/latex]; that is, the equation [latex]10= \dfrac{5}{12}(24)[/latex].
Our new definitions for the trig ratios work just as well for obtuse angles, even though [latex]\theta[/latex] is not technically “inside” a triangle, because we use the coordinates of [latex]P[/latex] instead of the sides of a triangle to compute the ratios.
Notice first of all that because [latex]x[/latex]-coordinates are negative in the second quadrant, the [latex]\cos[/latex] and tangent ratios are both negative for obtuse angles. For example, in the figure below, the point [latex](-4, 3)[/latex] lies on the terminal side of the angle [latex]\theta{.}[/latex] We see that [latex]r = \sqrt{(-4)^2 + 3^2} = 5{,}[/latex] so
Find the values of [latex]\cos \theta[/latex] and [latex]\tan \theta[/latex] if [latex]\theta[/latex] is an obtuse angle with [latex]\sin \theta = \dfrac{1}{3}{.}[/latex]
Solution
Because [latex]\theta[/latex] is obtuse, the terminal side of the angle lies in the second quadrant, as shown in the figure below. Because [latex]\sin \theta = \dfrac{1}{3}{,}[/latex] we know that [latex]\dfrac{y}{r} = \dfrac{1}{3}{,}[/latex] so we can choose a point [latex]P[/latex] with [latex]y=1[/latex] and [latex]r=3{.}[/latex] To find [latex]\cos \theta[/latex] and [latex]\tan \theta[/latex], we need to know the value of [latex]x{.}[/latex] From the Pythagorean theorem,
Remember that [latex]x[/latex] is negative in the second quadrant! Thus
[latex]\cos \theta = \dfrac{x}{r} = \dfrac{-\sqrt{8}}{3}[/latex] and [latex]\tan \theta = \dfrac{y}{x} = \dfrac{-1}{\sqrt{8}}[/latex].
Checkpoint 3.5.
Sketch an obtuse angle [latex]\theta[/latex] whose cosine is [latex]\dfrac{-8}{17}{.}[/latex]
Find the sin and the tangent of [latex]\theta{.}[/latex]
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 3.6.
Find the [latex]\sin[/latex] and [latex]\cos[/latex] of [latex]130°{.}[/latex] Compare to the [latex]\sin[/latex] and [latex]\cos[/latex] of [latex]50°{.}[/latex]
Solution
Using a calculator and rounding the values to four places, we find
[latex]\sin 130° = 0.7660[/latex] and [latex]\cos 130° = -0.6428[/latex]
[latex]\sin 50° = 0.7660[/latex] and [latex]\cos 50° = 0.6428[/latex]
We see that [latex]\sin 130° = \sin 50°[/latex] and [latex]\cos 130° = -\cos 50°{.}[/latex] This result should not be surprising when we look at both angles in standard position, as shown below.
The angles [latex]50°[/latex] and [latex]130°[/latex] are supplementary. The right triangles formed by choosing the points [latex](x,y)[/latex] and [latex](-x,y)[/latex] on their terminal sides are congruent triangles.
Consequently, the trigonometric ratios for [latex]50°[/latex] and for [latex]130°[/latex] are equal, except that the cosine of [latex]130°[/latex] is negative.
Checkpoint 3.7.
Use your calculator to fill in the table. Round to four decimal places.
[latex]\theta[/latex]
[latex]\cos \theta[/latex]
[latex]\sin \theta~~~~[/latex]
[latex]180° - \theta[/latex]
[latex]\cos (180° - \theta)[/latex]
[latex]\sin (180° - \theta)[/latex]
[latex]10 °[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]20 °[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]30 °[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]40 °[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]50 °[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]60 °[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]70 °[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]80 °[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
[latex]~[/latex]
Solution
[latex]\theta[/latex]
[latex]\cos \theta[/latex]
[latex]\sin \theta~~~~[/latex]
[latex]180° - \theta[/latex]
[latex]\cos (180° - \theta)[/latex]
[latex]\sin (180° - \theta)[/latex]
[latex]10 °[/latex]
[latex]0.9848[/latex]
[latex]0.1736[/latex]
[latex]170°[/latex]
[latex]-0.9848[/latex]
[latex]0.1736[/latex]
[latex]20 °[/latex]
[latex]0.9397[/latex]
[latex]0.3420[/latex]
[latex]160°[/latex]
[latex]-0.9397[/latex]
[latex]0.3420[/latex]
[latex]30 °[/latex]
[latex]0.8660[/latex]
[latex]0.5[/latex]
[latex]150°[/latex]
[latex]0.8660[/latex]
[latex]-0.5[/latex]
[latex]40 °[/latex]
[latex]0.7660[/latex]
[latex]0.6428[/latex]
[latex]140°[/latex]
[latex]-0.7660[/latex]
[latex]0.6428[/latex]
[latex]50 °[/latex]
[latex]0.6428[/latex]
[latex]0.7660[/latex]
[latex]130°[/latex]
[latex]-0.6428[/latex]
[latex]0.7660[/latex]
[latex]60 °[/latex]
[latex]0.5[/latex]
[latex]0.8660[/latex]
[latex]120°[/latex]
[latex]-0.5[/latex]
[latex]0.8660[/latex]
[latex]70 °[/latex]
[latex]0.3420[/latex]
[latex]0.9397[/latex]
[latex]110°[/latex]
[latex]-0.9397[/latex]
[latex]0.3420[/latex]
[latex]80 °[/latex]
[latex]0.1736[/latex]
[latex]0.9848[/latex]
[latex]100°[/latex]
[latex]-0.9848[/latex]
[latex]0.1736[/latex]
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 [latex]\theta{.}[/latex]
Because of these relationships, there are always two (supplementary) angles between [latex]0°[/latex] and [latex]180°[/latex] that have the same [latex]\sin[/latex]. Your calculator will only tell you one of them, so you have to be able to find the other one on your own! Fortunately, this is not difficult.
Example 3.9.
Find two different angles [latex]\theta{,}[/latex] rounded to the nearest [latex]0.1 °{,}[/latex] that satisfy [latex]\sin \theta = 0.25{.}[/latex]
Solution
To find an angle with [latex]\sin \theta = 0.25{,}[/latex] we calculate [latex]\theta = \sin^{-1}(0.25){.}[/latex] With the calculator in ° mode, we press 2nd sin 0.25 ENTER
to find that one angle is [latex]\theta \approx 14.5 °{.}[/latex] 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 [latex]\sin[/latex] is [latex]0.25{.}[/latex] To see the second angle, we draw a congruent triangle in the second quadrant as shown.
The supplement of [latex]14.5 °[/latex]—namely, [latex]\theta = 180° - 14.5 ° = 165.5°[/latex]—is the obtuse angle we need. Notice that [latex]\dfrac{y}{r} = 0.25[/latex] for both triangles, so [latex]\sin \theta = 0.25[/latex] for both angles.
Checkpoint 3.10.
Find two different angles [latex]\theta[/latex] that satisfy [latex]\sin \theta = 0.5{.}[/latex]
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.
Example 3.11.
Find the angle shown at right.
Solution
Using [latex]x=-3[/latex] and [latex]y=4{,}[/latex] we find [latex]r=\sqrt{3^2 + 4^2} = \sqrt{25} = 5[/latex] so [latex]\cos \theta = \dfrac{x}{r} = \dfrac{-3}{5}{,}[/latex] and [latex]\theta = \cos^{-1}\left( \dfrac{-3}{5}\right).[/latex] We can enter 2nd cos-3/5 ENTER to see that [latex]\theta \approx 126.9 °{.}[/latex]
Caution 3.12.
In the previous example, you might notice that [latex]\tan \theta = \dfrac{-4}{3}[/latex] and try to find by calculating [latex]\tan^{-1}(\dfrac{-4}{3}){.}[/latex] However, if we press
2nd TAN 4 ÷ 3 ENTER
the calculator returns an angle of [latex]\theta \approx -53.1 °{.}[/latex] It is true that [latex]\tan (-53.1 °) = \dfrac{-4}{3}{,}[/latex] but this is not the obtuse angle we want.
We also know that [latex]\sin \theta = \dfrac{4}{5}{,}[/latex] and if we press
2nd sin 4 ÷ 5 ENTER
we get [latex]\theta \approx 53.1 °{.}[/latex]
This is the acute angle whose terminal side passes through the point [latex](3,4){,}[/latex] as shown in the figure above. The angle we want is its supplement, [latex]\theta \approx 180° - 53.1° = 126.9°{.}[/latex]
Checkpoint 3.13.
Find the cosine of an obtuse angle with [latex]\tan \theta = -2[/latex] .
Find the angle [latex]\theta[/latex] in part [latex]a[/latex].
Solution
[latex]\dfrac{-1}{\sqrt{5}}[/latex]
[latex]\theta \approx 116.565°[/latex]
Supplements of the Special Angles
In Chapter 2 we learned that the angles [latex]30°, 45°[/latex], and [latex]60°[/latex] 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 3.14.
Find exact values for the trigonometric ratios of [latex]135 °{.}[/latex]
Solution
We sketch an angle of [latex]\theta = 135°[/latex] in standard position, as shown below. The terminal side is in the second quadrant, and it makes an acute angle of [latex]45°[/latex] with the negative [latex]x[/latex]-axis and passes through the point [latex](-1,1)[/latex]. Thus, [latex]~r=\sqrt{(-1)^2 +1^2} = \sqrt{2}~{,}[/latex] and we calculate
Find exact values for the trigonometric ratios of [latex]120°[/latex] and [latex]150°{.}[/latex]
Solution
[latex]\theta[/latex]
[latex]\cos \theta[/latex]
[latex]\sin \theta[/latex]
[latex]\tan \theta[/latex]
[latex]120°[/latex]
[latex]\dfrac{-1}{2}[/latex]
[latex]\dfrac{\sqrt{3}}{2}[/latex]
[latex]-\sqrt{3}[/latex]
[latex]150°[/latex]
[latex]\dfrac{-\sqrt{3}}{2}[/latex]
[latex]\dfrac{1}{2}[/latex]
[latex]\dfrac{-1}{\sqrt{3}}[/latex]
We can also find the trig ratios for the quadrantal angles. These are the angles, including [latex]0°{,}[/latex] [latex]90°[/latex], and [latex]180°{,}[/latex] whose terminal sides lie on one of the axes.
Example 3.16.
Find exact values for the trigonometric ratios of [latex]90°{.}[/latex]
Solution
The terminal side of a [latex]90°[/latex] angle in standard position is the positive [latex]y[/latex]-axis. If we take the point [latex]P(0,1)[/latex] on the terminal side, as shown at right, then [latex]x=0[/latex] and [latex]y=1{.}[/latex] Although we don’t have a triangle, we can still calculate a value for [latex]r{,}[/latex] the distance from the origin to [latex]P{.}[/latex]
[latex]r = \sqrt{0^2 + 1^2} = 1[/latex]
Our coordinate definitions for the trig ratios give us
[latex]\cos 90° = \dfrac{x}{r} = \dfrac{0}{1}[/latex] and [latex]\sin 90° = \dfrac{y}{r} = \dfrac{1}{1}[/latex]
so [latex]\cos 90° = 0[/latex] and [latex]\sin 90° = 1.[/latex] Also, [latex]\tan 90° = \dfrac{y}{x} = \dfrac{1}{0}~,[/latex] so [latex]\tan 90°[/latex] is undefined.
Checkpoint 3.17.
Find exact values for the trigonometric ratios of [latex]180°{.}[/latex]
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,
[latex]A = \dfrac{1}{2}bh{,}[/latex]
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 [latex]a[/latex] and [latex]b{,}[/latex] and the included angle, [latex]\theta{.}[/latex] The figure below shows three possibilities depending on whether the angle [latex]\theta[/latex] is acute, obtuse, or [latex]90°{.}[/latex]
In each case, [latex]b[/latex] is the base of the triangle, and its altitude is [latex]h{.}[/latex] Our task is to find an expression for [latex]h[/latex] in terms of the quantities we know: [latex]a{,}[/latex] [latex]b{,}[/latex] and [latex]\theta{.}[/latex] You should check that in all three triangles
[latex]\sin \theta = \dfrac{h}{a}[/latex]
Solving for [latex]h[/latex] gives us [latex]h = a \sin \theta{.}[/latex] Finally, we substitute this expression for [latex]h[/latex] into our old formula for the area to get
[latex]A = \dfrac{1}{2}b {h} = \dfrac{1}{2}b {a \sin \theta}[/latex]
Area of a Triangle.
Area of a Triangle.
If a triangle has sides of length [latex]a[/latex] and [latex]b{,}[/latex] and the angle between those two sides is [latex]\theta{,}[/latex] then the area of the triangle is given by
[latex]{A = \dfrac{1}{2} ab \sin \theta}[/latex]
Example 3.18.
Example 3.18.
Find the area of lot 86.
Solution
For the triangle in the lower portion of lot [latex]86[/latex], [latex]a = 120.3{,}[/latex] [latex]b = 141{,}[/latex] and [latex]\theta = 95°{.}[/latex] The area of that portion is
[latex]{\text{First Area}}= \dfrac{1}{2}absin \theta = \dfrac{1}{2} (120.3)(141)\sin 95° \approx 8448.88[/latex]
For the triangle in the upper portion of the lot, [latex]a = 161{,}[/latex] [latex]b = 114.8{,}[/latex] and [latex]\theta = 86.1°{.}[/latex] The area of that portion is
[latex]{\text{Second Area}}= \dfrac{1}{2}absin \theta = \dfrac{1}{2} (161)(114.8) \sin 86.1° \approx 9220.00[/latex]
The total area of the lot is the sum of the areas of the triangles
[latex]\text{Total area} = \text{First Area} + \text{Second Area} \approx 17668.88[/latex]
Lot [latex]86[/latex] has an area of approximately [latex]17,669[/latex] square feet.
Caution 3.19.
The formula [latex]~A= \dfrac{1}{2}absin \theta~[/latex] does not mean that we always use the sides labeled [latex]a[/latex] and [latex]b[/latex] to find the area of a triangle.
In this formula, the variables [latex]a[/latex] and [latex]b[/latex] represent the lengths of the sides that include the known angle. For example, the area of the triangle at right is given by [latex]A= \dfrac{1}{2}(5c)\sin \phi{.}[/latex]
Checkpoint 3.20.
A triangle has sides of length 6 and 7, and the angle between those sides is [latex]150°{.}[/latex] Find the area of the triangle.
Solution
[latex]\dfrac{21}{2}[/latex]
Review the following skills you will need for this section.
We put an angle in standard position by placing its vertex at the origin and the initial side on the positive [latex]x[/latex]-axis.
Coordinate Definitions of the Trigonometric Ratios.
[latex]\cos \theta = \dfrac{x}{r}[/latex]
[latex]\sin \theta = \dfrac{y}{r}[/latex]
[latex]tan \theta = \dfrac{y}{x}[/latex]
Trigonometric Ratios for Supplementary Angles.
[latex]\cos(180° - \theta) = -\cos \theta[/latex]
[latex]\sin(180° - \theta) = \sin \theta[/latex]
[latex]tan(180° - \theta) = -\tan \theta[/latex]
There are always two [latex]supplementary[/latex] angles between [latex]0°[/latex] and [latex]180°[/latex] that have the same sines. Your calculator will only tell you one of them.
Area of a Triangle.
If a triangle has sides of length [latex]a[/latex] and [latex]b{,}[/latex] and the angle between those two sides is [latex]\theta{,}[/latex] then the area of the triangle is given by
[latex]A = \dfrac{1}{2} ab \sin \theta[/latex]
Study Questions
Delbert says that [latex]\sin \theta = \dfrac{4}{7}[/latex] 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 [latex]-x[/latex].
Why are the \sins of supplementary angles equal, but the \cosines are not? What about the tangents of supplementary angles?
Use your calculator to evaluate [latex]\sin 118°{,}[/latex] then evaluate [latex]\sin^{-1} {ANS}[/latex] . Explain the result.
Write an expression for the area of the triangle.
definition
An oblique triangle is a triangle that does not have a right angle. In an oblique triangle, all three angles are acute (less than 90 degrees) or one angle is obtuse (greater than 90 degrees). Oblique triangles are sometimes also referred to as non-right triangles.
An angle is in standard position if its vertex is located at the origin and one ray is on the positive x-axis. The ray on the x-axis is called the initial side and the other ray is called the terminal side.
An angle is in standard position if its vertex is located at the origin and one ray is on the positive x-axis. The ray on the x-axis is called the initial side and the other ray is called the terminal side.
An angle is in standard position if its vertex is located at the origin and one ray is on the positive x-axis. The ray on the x-axis is called the initial side and the other ray is called the terminal side.
A quadrantal angle is an angle in the standard position and has a measure which is a multiple of [latex]90^o[/latex] or [latex]\frac{\pi}{2}[/latex] radians. It is an angle in standard position whose terminal ray lies along one of the axes.
A quadrilateral is a four-sided polygon with straight sides and four vertices.
In mathematics, an identity is an equation that is true for all values of the variables involved. In other words, it is a statement that holds true regardless of the values of the variables, and does not need to be proved or solved for a specific set of values. Examples of identity sin²θ + cos²θ = 1.