In this chapter, we showcase how the the tools we’ve developed in Chapters FoundationsofTrigonometry and AnalyticalTrigonometry can be applied to Geometry. Our first two sections focus specifically on solving oblique (non-right) Triangles.

Our first example reviews the basics of right triangle trigonometry. The reader is referred to Section AppRightTrig for more details and practice with these concepts.

A few remarks about Example righttrianglereviewex are in order. First, we adhere to the convention that a lower case Greek letter denotes an angle (as well as the measure of said angle) and the corresponding lowercase English letter represents the side (as well as the length of said side) opposite that angle.

More specifically, \(a\) is the side opposite \(\alpha \), \(b\) is the side opposite \(\beta \) and \(c\) is the side opposite \(\gamma \). Taken together, the pairs \((\alpha , a)\), \((\beta , b)\) and \((\gamma , c)\) are called angle-side opposite pairs.

Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it minimizes the chances of propagated error.

Third, since many of the applications which require solving triangles ‘in the wild’ rely on degree measure, we shall adopt this convention for the time being.

The Pythagorean Theorem along with Definition righttrianglesinecosinetangent allow us to easily handle any given right triangle problem, but what if the triangle isn’t a right triangle? In certain cases, we can use the Law of Sines.

The proof of the Law of Sines can be broken into three cases, and, as we’ll see, ultimately relies on what we know about right triangles.

For our first case, consider the triangle \(\triangle ABC\) below, all of whose angles are acute, with angle-side opposite pairs \((\alpha , a)\), \((\beta , b)\) and \((\gamma , c)\).

If we drop an altitude from vertex \(B\), we divide the triangle into two right triangles: \(\triangle ABQ\) and \(\triangle BCQ\).

If we call the length of the altitude \(h\) (for height), we get from Definition righttrianglesinecosinetangent that \(\sin (\alpha ) = \frac {h}{c}\) and \(\sin (\gamma ) = \frac {h}{a}\) so that \(h = c\sin (\alpha ) = a \sin (\gamma )\). Rearranging this last equation, we get \(\frac {\sin (\alpha )}{a} = \frac {\sin (\gamma )}{c}\).

Dropping an altitude from vertex \(A\), we can proceed as above using the triangles \(\triangle ABQ\) and \(\triangle ACQ\). We find that \(\frac {\sin (\beta )}{b} = \frac {\sin (\gamma )}{c}\), so we have shown \(\frac {\sin (\alpha )}{a} = \frac {\sin (\beta )}{b} = \frac {\sin (\gamma )}{c}\) as required.

For our next case consider the triangle \(\triangle ABC\) below with obtuse angle \(\alpha \).

Extending an altitude from vertex \(A\) gives two right triangles, as in the previous case: \(\triangle ABQ\) and \(\triangle ACQ\).

Proceeding as before, we get \(h = b \sin (\gamma )\) and \(h = c \sin (\beta )\) so that \(\frac {\sin (\beta )}{b} = \frac {\sin (\gamma )}{c}\).

Dropping an altitude from vertex B also generates two right triangles, \(\triangle ABQ\) and \(\triangle BCQ\).

We see \(\sin (\alpha ') = \frac {h'}{c}\) so that \(h' = c \sin (\alpha ')\). Since \(\alpha ' = 180^{\circ } - \alpha \), \(\sin (\alpha ') = \sin (\alpha )\), so \(h' = c\sin (\alpha )\).

Proceeding to \(\triangle BCQ\), we get \(\sin (\gamma ) = \frac {h'}{a}\) so \(h' = a \sin (\gamma )\).

As before, we get \(\frac {\sin (\gamma )}{c} = \frac {\sin (\alpha )}{a}\), so \( \frac {\sin (\alpha )}{a} = \frac {\sin (\beta )}{b} = \frac {\sin (\gamma )}{c}\) in this case, too.

The remaining case is when \(\triangle ABC\) is a right triangle. In this case, the Law of Sines reduces to the formulas given in Definition righttrianglesinecosinetangent and is left to the reader.

In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of the power, and the pitfalls, of the Law of Sines.

Some remarks about Example losex are in order. First note that if we are given the measures of two of the angles in a triangle, say \(\alpha \) and \(\beta \), the measure of the third angle \(\gamma \) is uniquely determined using the equation \(\gamma = 180^{\circ } - \alpha - \beta \). Knowing the measures of all three angles of a triangle completely determines the triangle’s shape.

If in addition we are given the length of one of the sides of the triangle, we can then use the Law of Sines to find the lengths of the remaining two sides to determine the size of the triangle. Such is the case in numbers losaas and losasa above.

In number losaas, the given side is adjacent to just one of the angles – this is called the ‘Angle-Angle-Side’ (AAS) case. In number losasa, the given side is adjacent to both angles which means we are in the so-called ‘Angle-Side-Angle’ (ASA) case.

If, on the other hand, we are given the measure of just one of the angles in the triangle along with the length of two sides, only one of which is adjacent to the given angle, we are in the ‘Angle-Side-Side’ (ASS) case. Such was the case in numbers losnotriangleex, losrighttriangleex, lostwotriangleex, and losonetriangleex above.

In number losnotriangleex, the length of the one given side \(a\) was too short to even form a triangle; in number losrighttriangleex, the length of \(a\) was just long enough to form a right triangle; in lostwotriangleex, \(a\) was long enough, but not too long, so that two triangles were possible; and in number losonetriangleex, side \(a\) was long enough to form a triangle but too long to swing back and form two. These four cases exemplify all of the possibilities in the Angle-Side-Side case which are summarized in the following theorem.

Theorem ASScase is proved on a case-by-case basis. If \(a < h\), then \(a < c\sin (\alpha )\). If a triangle were to exist, the Law of Sines would have \(\frac {\sin (\gamma )}{c} = \frac {\sin (\alpha )}{a}\) so that \(\sin (\gamma ) = \frac {c \sin (\alpha )}{a} > \frac {a}{a} = 1\), which is impossible.

In the figure below on the left, we see geometrically why this is the case. Simply put, if \(a < h\) the side \(a\) is too short to connect to form a triangle.

This means if \(a \geq h\), we are always guaranteed to have at least one triangle, and the remaining parts of the theorem tell us what kind and how many triangles to expect in each case.

If \(a = h\), then \(a = c\sin (\alpha )\) and the Law of Sines gives \(\frac {\sin (\alpha )}{a} = \frac {\sin (\gamma )}{c}\) so that \(\sin (\gamma ) = \frac {c \sin (\alpha )}{a} = \frac {a}{a} = 1\). Here, \(\gamma = 90^{\circ }\) as required. This situation is sketched below on the right.

Moving along, now suppose \(h < a < c\). As before, the Law of Sines gives \(\sin (\gamma ) = \frac {c \sin (\alpha )}{a}\).

Since \(h < a\), \(c \sin (\alpha ) < a\) or \(\frac {c\sin (\alpha )}{a} < 1\) which means there are two solutions to \(\sin (\gamma ) = \frac {c \sin (\alpha )}{a}\): an acute angle which we’ll call \(\gamma _0\), and its supplement, \(180^{\circ } - \gamma _0\).

Our job now is to argue that each of these angles ‘fit’ into a triangle with \(\alpha \). Since \((\alpha , a)\) and \((\gamma _0,c)\) are angle-side opposite pairs, the assumption \(c > a\) in this case gives us \(\gamma _0 > \alpha \). Since \(\gamma _0\) is acute, we must have that \(\alpha \) is acute as well. This means one triangle can contain both \(\alpha \) and \(\gamma _0\), giving us one of the triangles promised in the theorem.

If we manipulate the inequality \(\gamma _0 > \alpha \) a bit, we have \(180^{\circ } - \gamma _0 < 180^{\circ } - \alpha \). Adding \(\alpha \) to both sides gives \(\left (180^{\circ } - \gamma _0\right ) + \alpha < 180^{\circ }\). This proves a triangle can contain both of the angles \(\alpha \) and \(\left (180^{\circ } - \gamma _0\right )\), giving us the second triangle predicted in the theorem. We sketch the two triangle case below on the left.

To prove the last case in the theorem, we assume \(a \geq c\). Then \(\alpha \geq \gamma \), which forces \(\gamma \) to be an acute angle. Hence, we get only one triangle in this case, completing the proof.

One last comment regarding the Angle-Side-Side case: if you are given an obtuse angle to begin with then it is impossible to have the two triangle case. Think about this before reading further.

In many of the derivations and arguments in this section, we used the height of a given triangle, \(h\), as an intermediate variable to prove equivalences. Since the height of a triangle can be used to determine the area enclosed by said triangle, we can use the methods in this section to reformulate area in terms of side lengths and sines of angles. We state the following theorem and leave its proof as an exercise.

1 Bearings

Our last example of the section uses the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass.

The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must first understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose.

For example, N\(40^{\circ }\)E (read “\(40^{\circ }\) east of north”) is a bearing which is rotated clockwise \(40^{\circ }\) from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of \(\theta = 50^{\circ }\).

Similarly, S\(50^{\circ }\)W would point into Quadrant III along the terminal side of \(\theta = 220^{\circ }\) because we started out pointing due south (along \(\theta = 270^{\circ }\)) and rotated clockwise \(50^{\circ }\) back to \(220^{\circ }\).

Counter-clockwise rotations would be found in the bearings N\(60^{\circ }\)W (which is on the terminal side of \(\theta = 150^{\circ }\)) and S\(27^{\circ }\)E (which lies along the terminal side of \(\theta = 297^{\circ }\)).

These four bearings are sketched in the plane below.

The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as ‘due north’, ‘due south’, ‘due east’ and ‘due west’, respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction.

We make good use of bearings and the Law of Sines in our next example.