4.1: Right triangles
 Page ID
 89299
 How can we view \(\cos(\theta)\) and \(\sin(\theta)\) as side lengths in right triangles with hypotenuse \(1\text{?}\)
 Why can both \(\cos(\theta)\) and \(\sin(\theta)\) be thought of as ratios of certain side lengths in any right triangle?
 What is the minimum amount of information we need about a right triangle in order to completely determine all of its sides and angles?
In Section 2.3, we defined the cosine and sine functions as the functions that track the location of a point traversing the unit circle counterclockwise from \((1,0)\text{.}\) In particular, for a central angle of radian measure \(t\) that passes through the point \((1,0)\text{,}\) we define \(\cos(t)\) as the \(x\)coordinate of the point where the other side of the angle intersects the unit circle, and \(\sin(t)\) as the \(y\)coordinate of that same point, as pictured in Figure \(\PageIndex{1}\)
By changing our perspective slightly, we can see that it is equivalent to think of the values of the sine and cosine function as representing the lengths of legs in right triangles. Specifically, given a central angle^{ 1 } \(\theta\text{,}\) if we think of the right triangle with vertices \((\cos(\theta),0)\text{,}\) \((0,0)\text{,}\) and \((\cos(\theta), \sin(\theta))\text{,}\) then the length of the horizontal leg is \(\cos(\theta)\) and the length of the vertical leg is \(\sin(\theta)\text{,}\) as seen in Figure \(\PageIndex{2}\)
This right triangle perspective enables us to use the sine and cosine functions to determine missing information in certain right triangles. The field of mathematics that studies relationships among the angles and sides of triangles is called trigonometry. In addition, it's important to recall both the Pythagorean Theorem and the Fundamental Trigonometric Identity. The former states that in any right triangle with legs of length \(a\) and \(b\) and hypotenuse of length \(c\text{,}\) it follows \(a^2 + b^2 = c^2\text{.}\) The latter, which is a special case of the Pythagorean Theorem, says that for any angle \(\theta\text{,}\) \(\cos^2(\theta) + \sin^2(\theta) = 1\text{.}\)
For each of the following situations, sketch a right triangle that satisfies the given conditions, and then either determine the requested missing information in the triangle or explain why you don't have enough information to determine it. Assume that all angles are being considered in radian measure.
 The length of the other leg of a right triangle with hypotenuse of length \(1\) and one leg of length \(\frac{3}{5}\text{.}\)
 The lengths of the two legs in a right triangle with hypotenuse of length \(1\) where one of the nonright angles measures \(\frac{\pi}{3}\text{.}\)
 The length of the other leg of a right triangle with hypotenuse of length \(7\) and one leg of length \(6\text{.}\)
 The lengths of the two legs in a right triangle with hypotenuse \(5\) and where one of the nonright angles measures \(\frac{\pi}{4}\text{.}\)
 The length of the other leg of a right triangle with hypotenuse of length \(1\) and one leg of length \(\cos(0.7)\text{.}\)
 The measures of the two angles in a right triangle with hypotenuse of length \(1\) where the two legs have lengths \(\cos(1.1)\) and \(\sin(1.1)\text{,}\) respectively.
The geometry of triangles
In the study of functions, linear functions are the simplest of all and form a foundation for our understanding of functions that have other shapes. In the study of geometric shapes (polygons, circles, and more), the simplest figure of all is the triangle, and understanding triangles is foundational to understanding many other geometric ideas. To begin, we list some familiar and important facts about triangles.
 Any triangle has \(6\) important features: \(3\) sides and \(3\) angles.
 In any triangle in the Cartesian plane, the sum of the measures of the interior angles is \(\pi\) radians (or equivalently, \(180^\circ\)).
 In any triangle in the plane, knowing three of the six features of a triangle is often enough information to determine the missing three features.^{ 2 }
The situation is especially nice for right triangles, because then we only have five unknown features since one of the angles is \(\frac{\pi}{2}\) radians (or \(90^\circ\)), as demonstrated in Figure \(\PageIndex{3}\) If we know one of the two nonright angles, then we know the other as well. Moreover, if we know any two sides, we can immediately deduce the third, because of the Pythagorean Theorem. As we saw in Preview Activity \(\PageIndex{1}\), the cosine and sine functions offer additional help in determining missing information in right triangles. Indeed, while the functions \(\cos(t)\) and \(\sin(t)\) have many important applications in modeling periodic phenomena such as osciallating masses on springs, they also find powerful application in settings involving right triangles, such as in navigation and surveying.
Because we know the values of the cosine and sine functions from the unit circle, right triangles with hypotentuse \(1\) are the easiest ones in which to determine missing information. In addition, we can relate any other right triangle to a right triangle with hypotenuse \(1\) through the concept of similarity. Recall that two triangles are similar provided that one is a magnification of the other. More precisely, two triangles are similar whenever there is some constant \(k\) such that every side in one triangle is \(k\) times as long as the corresponding side in the other and the corresponding angles in the two triangles are equal. An important result from geometry tells us that if two triangles are known to have all three of their corresponding angles equal, then it follows that the two triangles are similar, and therefore their corresponding sides must be proportionate to one another.
Consider right triangle \(OPQ\) given in Figure \(\PageIndex{4}\), and assume that the length of the hypotenuse is \(OP = r\) for some constant \(r \gt 1\text{.}\) Let point \(M\) lie on \(\overline{OP}\) between \(O\) and \(P\) in such a way that \(OM = 1\text{,}\) and let point \(N\) lie on \(\overline{OQ}\) so that \(\angle ONM\) is a right angle, as pictured. In addition, assume that point \(O\) corresponds to \((0,0)\text{,}\) point \(Q\) to \((x,0)\text{,}\) and point \(P\) to \((x,y)\) so that \(OQ = x\) and \(PQ = y\text{.}\) Finally, let \(\theta\) be the measure of \(\angle POQ\text{.}\)
 Explain why \(\triangle OPQ\) and \(\triangle OMN\) are similar triangles.
 What is the value of the ratio \(\frac{OP}{OM}\text{?}\) What does this tell you about the ratios \(\frac{OQ}{ON}\) and \(\frac{PQ}{MN}\text{?}\)
 What is the value of \(ON\) in terms of \(\theta\text{?}\) What is the value of \(MN\) in terms of \(\theta\text{?}\)
 Use your conclusions in (b) and (c) to express the values of \(x\) and \(y\) in terms of \(r\) and \(\theta\text{.}\)
Ratios of sides in right triangles
A right triangle with a hypotenuse of length \(1\) can be viewed as lying in standard position in the unit circle, with one vertex at the origin and one leg along the positive \(x\)axis. If we let the angle formed by the hypotenuse and the horizontal leg have measure \(\theta\text{,}\) then the right triangle with hypotenuse \(1\) has horizontal leg of length \(\cos(\theta)\) and vertical leg of length \(\sin(\theta)\text{.}\) If we consider now consider a similar right triangle with hypotenuse of length \(r \ne 1\text{,}\) we can view that triangle as a magnification of a triangle with hypotenuse \(1\text{.}\) These observations, combined with our work in Activity \(\PageIndex{2}\), show us that the horizontal legs of the right triangle with hypotenuse \(r\) have measure \(r\cos(\theta)\) and \(r\sin(\theta)\text{,}\) as pictured in Figure \(\PageIndex{5}\)
From the similar triangles in Figure \(\PageIndex{5}\), we can make an important observation about ratios in right triangles. Because the triangles are similar, the ratios of corresponding sides must be equal, so if we consider the two hypotenuses and the two horizontal legs, we have
If we rearrange Equation (4.1.1) by dividing both sides by \(r\) and multiplying both sides by \(\cos(\theta)\text{,}\) we see that
From a geometric perspective, Equation (4.1.2) tells us that the ratio of the horizontal leg of a right triangle to the hypotenuse of the triangle is always the same (regardless of \(r\)) and that the value of that ratio is \(\cos(\theta)\text{,}\) where \(\theta\) is the angle adjacent to the horizontal leg. In an analogous way, the equation involving the hypotenuses and vertical legs of the similar triangles is
which can be rearranged to
Equation (4.1.4) shows that the ratio of the vertical leg of a right triangle to the hypotenuse of the triangle is always the same (regardless of \(r\)) and that the value of that ratio is \(\sin(\theta)\text{,}\) where \(\theta\) is the angle opposite the vertical leg. We summarize these recent observations as follows.
In a right triangle where one of the nonright angles is \(\theta\text{,}\) and “adj” denotes the length of the leg adjacent to \(\theta\text{,}\) “opp” the length the side opposite \(\theta\text{,}\) and “hyp” the length of the hypotenuse,
\[ \cos(\theta) = \frac{\text{adj}}{\text{hyp}} \text{ and } \sin(\theta) = \frac{\text{opp}}{\text{hyp}}\text{.} \nonumber \]
In each of the following scenarios involving a right triangle, determine the exact values of as many of the remaining side lengths and angle measures (in radians) that you can. If there are quantities that you cannot determine, explain why. For every prompt, draw a labeled diagram of the situation.
 A right triangle with hypotenuse \(7\) and one nonright angle of measure \(\frac{\pi}{7}\text{.}\)
 A right triangle with nonright angle \(\alpha\) that satisfies \(\sin(\alpha) = \frac{3}{5}\text{.}\)
 A right triangle where one of the nonright angles is \(\beta = 1.2\) and the hypotenuse has length \(2.7\text{.}\)
 A right triangle with hypotenuse \(13\) and one leg of length \(6.5\text{.}\)
 A right triangle with legs of length \(5\) and \(12\text{.}\)
 A right triangle where one of the nonright angles is \(\beta = \frac{\pi}{5}\) and the leg opposite this angle has length \(4\text{.}\)
Using a ratio involving sine and cosine
In Activity \(\PageIndex{3}\), we found that in many cases where we have a right triangle, knowing two additional pieces of information enables us to find the remaining three unknown quantities in the triangle. At this point in our studies, the following general principles hold.
In any right triangle,
 if we know one of the nonright angles and the length of the hypotenuse, we can find both the remaining nonright angle and the lengths of the two legs;
 if we know the length of two sides of the triangle, then we can find the length of the other side;
 if we know the measure of one nonright angle, then we can find the measure of the remaining angle.
In scenario (1.), all \(6\) features of the triangle are not only determined, but we are able to find their values. In (2.), the triangle is uniquely determined by the given information, but as in Activity 4.1.3 parts (d) and (e), while we know the values of the sine and cosine of the angles in the triangle, we haven't yet developed a way to determine the measures of those angles. Finally, in scenario (3.), the triangle is not uniquely determined, since any magnified version of the triangle will have the same three angles as the given one, and thus we need more information to determine side length.
We will revisit scenario (2) in our future work. Now, however, we want to consider a situation that is similar to (1), but where it is one leg of the triangle instead of the hypotenuse that is known. We encountered this in Activity 4.1.3 part (f): a right triangle where one of the nonright angles is \(\beta = \frac{\pi}{5}\) and the leg opposite this angle has length \(4\text{.}\)
Consider a right triangle in which one of the nonright angles is \(\beta = \frac{\pi}{5}\) and the leg opposite \(\beta\) has length \(4\text{.}\)
Determine (both exactly and approximately) the measures of all of the remaining sides and angles in the triangle.
 Solution

From the fact that \(\beta = \frac{\pi}{5}\text{,}\) it follows that \(\alpha = \frac{\pi}{2}  \frac{\pi}{5} = \frac{3\pi}{10}\text{.}\) In addition, we know that
\[ \sin\left(\frac{\pi}{5}\right) = \frac{4}{h}\label{eqrightsinratio}\tag{4.1.5} \]
and\[ \cos\left(\frac{\pi}{5}\right) = \frac{x}{h}\label{eqrightcosratio}\tag{4.1.6} \]
Solving Equation (4.1.5) for \(h\text{,}\) we see that\[ h = \frac{4}{\sin\left(\frac{\pi}{5}\right)}\text{,}\label{eqrightsinratio2}\tag{4.1.7} \]
which is the exact numerical value of \(h\text{.}\) Substituting this result in Equation (4.1.6), solving for \(h\) we find that\[ \cos\left(\frac{\pi}{5}\right) = \frac{x}{\frac{4}{\sin\left(\frac{\pi}{5}\right)}}\text{.}\label{eqrightcosratio2}\tag{4.1.8} \]
Solving this equation for the single unknown \(x\) shows that\[ x = \frac{4 \cos\left(\frac{\pi}{5}\right)}{\sin\left(\frac{\pi}{5}\right)}\text{.} \nonumber \]
The approximate values of \(x\) and \(h\) are \(x \approx 5.506\) and \(h \approx 6.805\text{.}\)
Example \(\PageIndex{6}\) demonstrates that a ratio of values of the sine and cosine function can be needed in order to determine the value of one of the missing sides of a right triangle, and also that we may need to work with two unknown quantities simultaneously in order to determine both of their values.
We want to determine the distance between two points \(A\) and \(B\) that are directly across from one another on opposite sides of a river, as pictured in Figure \(\PageIndex{8}\) We mark the locations of those points and walk \(50\) meters downstream from \(B\) to point \(P\) and use a sextant to measure \(\angle BPA\text{.}\) If the measure of \(\angle BPA\) is \(56.4^{\circ}\text{,}\) how wide is the river? What other information about the situation can you determine?
Summary
 In a right triangle with hypotenuse \(1\text{,}\) we can view \(\cos(\theta)\) as the length of the leg adjacent to \(\theta\) and \(\sin(\theta)\) as the length of the leg opposite \(\theta\text{,}\) as seen in Figure 4.1.2. This is simply a change in perspective achieved by focusing on the triangle as opposed to the unit circle.
 Because a right triangle with hypotenuse of length \(r\) can be thought of as a scaled version of a right triangle with hypotenuse of length \(1\text{,}\) we can conclude that in a right triangle with hypotenuse of length \(r\text{,}\) the leg adjacent to angle \(\theta\) has length \(r\cos(\theta)\text{,}\) and the leg opposite \(\theta\) has length \(r\sin(\theta)\text{,}\) as seen in Figure 4.1.5. Moreover, in any right triangle with angle \(\theta\text{,}\) we know that
\[ \cos(\theta) = \frac{\text{adj}}{\text{hyp}} \text{ and } \sin(\theta) = \frac{\text{opp}}{\text{hyp}}\text{.} \nonumber \]
 In a right triangle, there are five additional characteristics: the measures of the two nonright angles and the lengths of the three sides. In general, if we know one of those two angles and one of the three sides, we can determine all of the remaining pieces.
Exercises
A person standing \(50\) feet away from a streetlight observes that they cast a shadow that is \(14\) feet long. If a ray of light from the streetlight to the tip of the person's shadow forms an angle of \(27.5^\circ\) with the ground, how tall is the person and how tall is the streetlight? What other information about the situation can you determine?
A person watching a rocket launch uses a laser rangefinder to measure the distance from themselves to the rocket. The rangefinder also reports the angle at which the finder is being elevated from horizontal. At a certain instant, the rangefinder reports that it is elevated at an angle of \(17.4^\circ\) from horizontal and that the distance to the rocket is \(1650\) meters. How high off the ground is the rocket? Assuming a straightline vertical path for the rocket that is perpendicular to the earth, how far away was the rocket from the rangefinder at the moment it was launched?
A trough is constructed by bending a \(4' \times 24'\) rectangular sheet of metal. Two symmetric folds \(2\) feet apart are made parallel to the longest side of the rectangle so that the trough has crosssections in the shape of a trapezoid, as pictured in Figure \(\PageIndex{9}\) Determine a formula for \(V(\theta)\text{,}\) the volume of the trough as a function of \(\theta\text{.}\)
 Hint

The volume of the trough is the area of a crosssection times the length of the trough.