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.
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.
Solution.
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.
Answer.
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]2ndTAN 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]
Solution.
Enter 2ndSIN 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.
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]
Answer.
[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.
Solution.
A sketch of the hill is shown at right. (Recall that [latex]1[/latex] mile [latex]= 5280[/latex] feet.)
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.
Answer.
[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]
Solution.
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]
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.
Trigonometric Ratios for the Special Angles
Angle
Sine
Cosine
Tangent
[latex]30{^o}[/latex]
[latex]\dfrac{1}{2} = 0.5[/latex]
[latex]\dfrac{\sqrt{3}}{2} \approx 0.8660[/latex]
[latex]\dfrac{1}{\sqrt{3}} \approx 0.5774[/latex]
[latex]45{^o}[/latex]
[latex]\dfrac{1}{\sqrt{2}} \approx 0.7071[/latex]
[latex]\dfrac{1}{\sqrt{2}} \approx 0.7071[/latex]
[latex]1[/latex]
[latex]60{^o}[/latex]
[latex]\dfrac{\sqrt{3}}{2} \approx 0.8660[/latex]
[latex]\dfrac{1}{2} = 0.5[/latex]
[latex]\sqrt{3} \approx 1.732[/latex]
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.
Note 2.42.
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.
Example 2.43.
If [latex]\cos \alpha \gt \dfrac{\sqrt{3}}{2}\text{,}[/latex] what can we say about [latex]\alpha\text{?}[/latex]
Solution.
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]
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.
The exact values of trigonometric ratios of the special angles should be memorized.
Angle
Sine
Cosine
Tangent
[latex]30{^o}[/latex]
[latex]\dfrac{1}{2} = 0.5[/latex]
[latex]\dfrac{\sqrt{3}}{2} \approx 0.8660[/latex]
[latex]\dfrac{1}{\sqrt{3}} \approx 0.5774[/latex]
[latex]45{^o}[/latex]
[latex]\dfrac{1}{\sqrt{2}} \approx 0.7071[/latex]
[latex]\dfrac{1}{\sqrt{2}} \approx 0.7071[/latex]
[latex]1[/latex]
[latex]60{^o}[/latex]
[latex]\dfrac{\sqrt{3}}{2} \approx 0.8660[/latex]
[latex]\dfrac{1}{2} = 0.5[/latex]
[latex]\sqrt{3} \approx 1.732[/latex]
You can remember the trig values for the special angles if you memorize two triangles:
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?
definition
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.
