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# 1.5: Triangles and Vectors

Difficulty Level: At Grade Created by: CK-12

## The Law of Cosines

Introduction

The definition of “oblique” (non-right) may be worth stressing briefly, as students may otherwise confuse it with “obtuse.” (Adding to the confusion is the fact that obtuse triangles are also oblique triangles, although not all oblique triangles are obtuse.)

Derive the Law of Cosines

The fourth line of the derivation here contains a clever trick that is worth explaining. The text explains why we can substitute \begin{align*}a \cos C\end{align*} for \begin{align*}x\end{align*}, but it may not be obvious why it is a good idea to do so. The reason is that it lets us express \begin{align*}c^2\end{align*} solely in terms of \begin{align*}a, b,\end{align*} and \begin{align*}C\end{align*}, so we now have a formula we can use for any triangle without needing to draw an altitude like BD again.

Side of an Oblique Triangle (given the other two sides)

“Note that the negative answer is thrown out as having no geometric meaning in this case” may bear explaining. The final step of example 1 involves taking the square root of both sides, which yields two possible answers, positive and negative. But we can safely disregard the negative answer because the number we are looking for represents the length of a line segment, which must be positive. This will be the case whenever we use the Law of Cosines.

Part 2 of the Real-World Application problem involves using the Law of Cosines differently, to find an angle when the sides are known instead of a side when the other two sides and an angle are known. This technique won’t actually be explained until the next section, so you may need to walk through this example very carefully, or just skip ahead to the next section and then come back to it.

Identify Accurate Drawings of General Triangles

A problem like Example 3 could of course be done faster if we were in a hurry; we could simply skip to part 2, since finding out what the angle should be would also tell us if the angle given was the correct one.

Points to Consider

1) If we apply the Law of Cosines to a right triangle, the term \begin{align*}2ab \cos C\end{align*} becomes zero because cos \begin{align*}90^\circ\end{align*} is zero. The Law of Cosines then reduces to the Pythagorean Theorem.

2) It is possible to solve the triangle completely through repeated applications of the Law of Cosines. Knowing two sides and the included angle lets us find the third side; then we can apply the theorem backwards, plugging in the three sides to find either of the missing angles; and then we can do this again to find the other missing angle, or simply find it by the Triangle Sum Theorem.

3) We cannot use the Law of Cosines if we only know one or no side lengths; if we know two side lengths but no angles; or if we know two side lengths and one angle, but the angle is not between the two sides. (It may seem that we could still apply the Law of Cosines in the latter case, but for some triangles it turns out that that yields two positive answers, and there is no way to tell which is correct. This is the Ambiguous Case described in a later lesson.)

4) Students may need to experiment a bit to answer this question. If they are stumped, remind them of the Triangle Inequality they may have learned in Geometry: the sum of any two sides of a triangle must be greater than the third side. Any set of three numbers such that one of them is greater than the sum of the other two is a set of numbers that do not form a triangle.

## Area of a Triangle

Derive Area \begin{align*}= \frac{1}{2} bc \sin A\end{align*}

The second line of the derivation here employs much the same trick as was used to derive the Law of Cosines in the previous lesson, and for similar reasons.

In estimating the cost of the pool cover here, we are simply calculating the exact cost by multiplying the precise area in square feet by \begin{align*}\35\end{align*}, for simplicity’s sake. For a small extra challenge, ask students to find the total cost if the price is a more realistic \begin{align*}\35\end{align*} per square foot or fraction thereof. (To solve this, they need to round up the area to the nearest whole number and then multiply by \begin{align*}\35\end{align*}.) For a bigger extra challenge, ask them to find the cost if the cover of the pool needs to be \begin{align*}1\end{align*} foot longer on each side than the pool itself. (To solve this, they would use the same formula as before, but \begin{align*}b\end{align*} and \begin{align*}c\end{align*} would be \begin{align*}25\end{align*} and \begin{align*}27\end{align*} instead of \begin{align*}24\end{align*} and \begin{align*}26\end{align*}.)

Find the Area Using Three Sides--SSS (side-side-side) Heron’s Formula

Observant students might worry that the terms (s-a), (s-b), and (s-c) in Heron’s Formula could end up being negative, which could make the whole expression under the square root sign negative and make it impossible to find the square root. However, inspecting the formula for s shows us that the term (s-a) can only be negative if a is greater than b+c (similar reasoning holds for (s-b) and (s-c)), and the Triangle Inequality (mentioned in the notes to the previous lesson) guarantees that this can never be the case.

Incidentally, calculators will be needed throughout this chapter, as we are no longer working with special angles whose trig ratios we know, or even angles we can look up in tables like the one in chapter 1.

Applications, Technological Tools

The sailboat diagram may be somewhat confusing. The jib sail is the gray shaded area on the left-hand side of the mast, and the line labeled \begin{align*}x\text{''}\end{align*} is the rope attaching the sail to the mast.

Points to Consider

1) The Triangle Inequality is the answer to this question as well. If any one side were greater than half the perimeter, it would be greater than the sum of the other two sides, which the Triangle Inequality tells us is impossible.

2) In these three cases, we actually don’t have enough information to solve the triangle or find its area using the techniques covered so far. We need to use the Law of Sines, which is the subject of the next lesson.

3) Applying Heron’s Formula in reverse, though tedious, will yield the third side if the first two sides and the area are known.

## The Law of Sines

Introduction

Note that in the diagram accompanying the Real-World Application problem, A represents Chicago and \begin{align*}C\end{align*} represents Buffalo; make sure students don’t get caught thinking \begin{align*}B\end{align*} is Buffalo and \begin{align*}C\end{align*} is Chicago.

The phrase “the side is not included” may also cause confusion. “Included” is being used here in the technical sense, so “not included” means that the side is not in between the two angles—even though the length of the side is clearly “included” in the set of information we are given.

Derive Two Forms of the Law of Sines

An interesting exercise is to apply the Law of Sines to a right triangle to verify its accuracy. If \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are the legs of the triangle and \begin{align*}c\end{align*} is the hypotenuse, then \begin{align*}\frac{a} {\sin A} = \frac{a} {\frac{a}{c}} = c\end{align*}. Similarly, \begin{align*}\frac{b} {\sin B} = \frac{b}{\frac{b}{c}} = c\end{align*}, and \begin{align*}\frac{c} {\sin C} = \frac{c}{1} = c\end{align*} (because \begin{align*}\sin 90^\circ = 1\end{align*}). As predicted, the ratios of all three sides to the sines of their opposite angles are equal.

AAS (Angle-Angle-Side)

An isosceles triangle appears in the Real-World Application problem here; we know that sides \begin{align*}BC\end{align*} and \begin{align*}DC\end{align*} are congruent because the angles across from them are congruent. A question for students: how could the Law of Sines have told us this if we didn’t already know it? (Hint: if \begin{align*}\frac{a}{\sin A} = \frac{b} {\sin B}\end{align*}, and \begin{align*}A = B\end{align*} (so \begin{align*}\sin A = \sin B\end{align*}), what does that tell us about \begin{align*}a\end{align*} and \begin{align*}b\end{align*}?)

ASA (Angle-Side-Angle)

The ASA case is very close to the AAS case; in fact, by using the Triangle Sum Theorem to find the third angle, we are really turning the ASA case into the AAS case, because we now know two angles and a side that is not between them.

For students who are developing headaches trying to memorize all these three-letter acronyms, there is a simpler way to figure out when it is possible to use the Law of Sines. If you know two angles of a triangle, you really know all three angles, so if you know any two angles and one side, you can use the Law of Sines to find the other two sides.

Applications

The diagram in situation \begin{align*}2\end{align*} is incompletely labeled : the top of the mountain should be marked \begin{align*}M\end{align*}, the right angle \begin{align*}N\end{align*}, and the \begin{align*}127^\circ\end{align*} angle \begin{align*}U\end{align*}. Side \begin{align*}u\end{align*} is then the same as side \begin{align*}x\end{align*}.

Points to Consider

1) We still can’t use the Law of Sines or Cosines if we don’t know any of a triangle’s side lengths, or if we only know one side and one angle. Also, we can’t use them if we only know two sides and an angle that is not between them.

2) Two angles can have the same sine if they are each other’s complements. This is why we can use the Law of Sines to solve for a side, but not for an angle; if we used it to solve for an angle, we might get two possible angles and not know which was correct.

3) With the Laws of Sines and Cosines together, we can solve any triangle if we know all three sides; two angles and one side; or two sides and the angle between them.

## The Ambiguous Case

Introduction

Note that the cases where the Law of Sines is useful are the cases where we know at least two angles. Thanks to the Triangle Sum Theorem, knowing two angles of a triangle really means we know all three angles, so the Law of Sines is useful when we know all the angles of a triangle (plus at least one side) and just need to find the remaining sides. It may therefore be useful to think “Use the Law of Sines to find the sides.” (Just remember that this little saying doesn’t always apply when we only know one angle—sometimes we need the Law of Cosines in that case.)

Possible Triangles Given SSA

In the case where \begin{align*}a < b\end{align*}, we can see why comparing a with \begin{align*}b \sin A\end{align*} tells us how many solutions there are if we think in terms of the Law of Sines. The Law of Sines tells us that \begin{align*}b \sin A = a \sin B\end{align*}, and that equation in turn tells us the following things:

1) If a is less than \begin{align*}b \sin A\end{align*}, \begin{align*}\sin B\end{align*} must be greater than 1 in order to make \begin{align*}a \sin B\end{align*} equal to \begin{align*}b \sin A\end{align*}. (If a by itself isn’t “big enough” to equal \begin{align*}b \sin A\end{align*}, it has to be multiplied by something greater than \begin{align*}1\end{align*} to make the whole expression big enough.) But there is no angle whose sine is greater than \begin{align*}1\end{align*}, so there is no solution.

2) Conversely, if a is greater than \begin{align*}b \sin A\end{align*}, \begin{align*}\sin B\end{align*} must be less than \begin{align*}1\end{align*} in order to make \begin{align*}a \sin B\end{align*} equal to \begin{align*}b \sin A\end{align*}. But for any given value of the sine function between \begin{align*}0\end{align*} and \begin{align*}1\end{align*}, there are two different angles between \begin{align*}0^\circ\end{align*} and \begin{align*}180^\circ\end{align*} that correspond to it—one acute angle and one obtuse angle, which are each other’s supplements. So there are two possible solutions.

3) Finally, if a is equal to \begin{align*}b \sin a\end{align*}, \begin{align*}\sin B\end{align*} must equal \begin{align*}1\end{align*} to keep the equation true—and there is only one angle between \begin{align*}0^\circ\end{align*} and \begin{align*}180^\circ\end{align*} whose sine is \begin{align*}1\end{align*}: a right angle. So there is one solution.

Points to Consider

1) One way to write the Law of Sines is \begin{align*}\frac{a}{b} = \frac{\sin A}{\sin B}\end{align*}. This tells us that if a is greater than \begin{align*}b, \sin A\end{align*} is also greater than \begin{align*}\sin B\end{align*}. The sine function is greater the closer an angle is to \begin{align*}90^\circ\end{align*}, so A must be closer to \begin{align*}90^\circ\end{align*} than \begin{align*}B\end{align*}.

• If \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are both less than \begin{align*}90^\circ\end{align*}, this means that \begin{align*}A\end{align*} is greater than \begin{align*}B\end{align*}.
• If \begin{align*}A\end{align*} is \begin{align*}90^\circ\end{align*} or greater, we already know \begin{align*}A\end{align*} is greater than \begin{align*}B\end{align*} because only one angle in a triangle can be \begin{align*}90^\circ\end{align*} or greater.
• If \begin{align*}B\end{align*} is \begin{align*}90^\circ\end{align*} or greater, then \begin{align*}A\end{align*} being closer to \begin{align*}90^\circ\end{align*} than \begin{align*}B\end{align*} means that \begin{align*}(90^\circ - A)\end{align*} must be less than \begin{align*}(B - 90^\circ)\end{align*}, which means \begin{align*}A + B > 180^\circ\end{align*}—but that’s not possible if \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are two angles of a triangle, so that means \begin{align*}B\end{align*} just can’t be \begin{align*}90^\circ\end{align*} or greater.

So, the Law of Sines tells us that if \begin{align*}a > b, A > B\end{align*} and (more importantly) \begin{align*}B < 90^\circ\end{align*}. This means that if we apply the Law of Sines when \begin{align*}a > b\end{align*} and get two possible values for angle \begin{align*}B\end{align*}, since one will be acute and one will be obtuse, we know that only the acute one can be correct.

(Applied more generally, the Law of Sines tells us that the biggest angle is opposite the biggest side and the smallest angle is opposite the smallest side. This is often useful to know.)

2) This was covered in the section above.

3) We can check whether each of the possible angles makes the total angle sum greater than \begin{align*}180^\circ\end{align*}. If it does, we throw it out; if not, it is a correct solution.

## General Solutions of Triangles

Summary of Triangle Techniques

Students should note that by applying the techniques in this table repeatedly, they can find all of the missing sides or angles in a triangle as long as they start out with enough information to find one of them.

Using the Law of Sines

In the Real-World Application problem, students are told to find the biggest angle first using the Law of Cosines in order to avoid dealing with the Ambiguous Case later when using the Law of Sines. This is because, as discussed in the notes to the previous lesson, only the biggest angle in a triangle can be obtuse, so finding the biggest angle first ensures that the other angles will be acute and will thus have only one possible value.

However, students should beware inaccurate diagrams! In a drawing that is not to scale (or one that might not be to scale), the angle that looks the biggest may not be the biggest. Instead of looking for what seems to be the biggest angle in the diagram given, if we know the exact side lengths we should make use of the fact that the biggest angle is opposite the longest side. (We can derive this fact from the Law of Sines, as described earlier.)

It may be best to skip the part that describes converting the angles to headings, as those headings can’t be found accurately without knowing which way is due north .

Points to Consider

1) It’s possible, but never really necessary, to use the Law of Sines before the Law of Cosines. There are times when we have enough information to use the Law of Sines but not the Law of Cosines, but in those cases, applying the Law of Sines always gives us enough information to finish solving the triangle without needing the Law of Cosines any more. Specifically, after applying the Law of Sines, no matter what information we started out with, we always end up knowing at least two sides and two angles, which means we really know two sides and three angles (by the Triangle Sum Theorem)—so one more application of the Law of Sines will give us the third side, and that’s all we have left to find.

2) If we know three sides and one angle, we could apply the Law of Sines, but we might then find ourselves in the Ambiguous Case. To avoid that, we might prefer to use the Law of Cosines. In any other case where both laws are applicable, though, the Law of Sines is generally preferable simply because it is easier.

3) In both cases where the Law of Cosines is applicable, we end up knowing all three sides and one angle after we apply it. Therefore, we can then switch to the Law of Sines (SSA case), although as just mentioned, we may prefer to stick with the Law of Cosines instead.

## Vectors

Introduction

In the definitions of displacement, velocity, and force, the phrase “in a certain direction” is the important part. Displacement without direction is simply distance; velocity without direction is simply speed; and there is no special term for force without direction. These definitions are taken from the study of physics.

Since distance \begin{align*}+\end{align*} direction \begin{align*}=\end{align*} displacement and speed \begin{align*}+\end{align*} direction \begin{align*}=\end{align*} velocity, we can say that distance is simply the magnitude of the vector that represents displacement, and speed is the magnitude of the vector that represents velocity.

Directed Line Segments, Equal Vectors, and Absolute Value

Drawing the vector in example 1 is probably a good idea, as is reviewing the distance formula. Drawing a right triangle with the given vector as the hypotenuse will remind students of how the distance formula is derived from the Pythagorean Theorem. When students use the distance formula throughout the rest of this chapter, drawing right triangles will help them to remember it, but it is also a good idea for them to try to learn the formula well enough to use it without consulting diagrams.

The direction of a vector, by the way, is defined as the angle made by the vector when it is placed in standard position.

You’ll need to skip over the section on subtraction until after you’ve covered the section on addition just below it.

Not only can we not use the parallelogram method to find the sum of a vector and itself, we also can’t use it to find the sum of a vector and its opposite (and in the latter case, the tip-to-tail method doesn’t work either.) However, the sum of a vector and its opposite is just \begin{align*}0\end{align*}, and the sum of a vector and itself is just a vector with twice the magnitude in the same direction.

Resultant of Two Displacements

When we return to the problem about the ship, you may need to remind students that the magnitude of each vector represents the speed the ship is traveling in that direction, and therefore that the total speed is represented by the magnitude of the resultant vector.

In the balloon example, angle \begin{align*}A\end{align*} is the “bottom corner” of the triangle in the diagram. There is an easier way to find the angle with the horizontal, though, and you might ask students what it is. (Answer: The “top right corner” of the triangle is the angle we are looking for, and its tangent is \begin{align*}\frac{13}{22}\end{align*}, from which we can calculate the angle directly.)

In case anyone is confused by part c of the “other things to consider” section, the phrase “(\begin{align*}22\mathrm{ft/second}\end{align*} times \begin{align*}14,400\;\mathrm{seconds}\end{align*} in two hours)” should read “four hours.” The numbers are still correct.

Points to Consider

When we add vectors using the triangle method, we know their lengths and can figure out the angle between them; hence we can use the Law of Cosines to figure out the magnitude and direction of the resultant vector.

This is a little less straightforward than it sounds, though, because we can’t just read off the angle measures easily. Consider the following example, where \begin{align*}A\end{align*} and \begin{align*}B\end{align*} represent the directional angles of the vectors a and b that are being added by the triangle method:

Unfortunately, \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are the angles the vectors make with a horizontal line, not the angles within the triangle which we need to know in order to use the Law of Cosines—so first we need to use geometry to determine the angles within the triangle. We can see that the largest angle is equal to \begin{align*}A\end{align*} (by alternate interior angles) plus the supplement of \begin{align*}B\end{align*}, and since we know sides \begin{align*}a\end{align*} and \begin{align*}b\end{align*}, we can use the Law of Cosines to find the magnitude of the resultant vector. Then the Law of Sines will tell us the other angles of the triangle, from which we can figure out angle \begin{align*}R\end{align*}, the direction angle of the resultant vector.

## Component Vectors

Vector Times a Scalar

Remind students here that \begin{align*}|\vec a|\end{align*} represents the magnitude of vector \begin{align*}\vec a\end{align*}.

It may be worth explaining why we can “scale” a vector up or down just by multiplying each of its coordinates by the same scalar. Referring back to the methods from the previous lesson will show why this works: if you take the formula for finding the magnitude of the original vector, and multiply all the coordinates by the constant \begin{align*}k\end{align*}, the result will be \begin{align*}k\end{align*} times the original magnitude; and if you use trig ratios to find the direction of the vector, those ratios will stay the same (and thus the direction stays the same) if you multiply all the coordinates by \begin{align*}k\end{align*}.

Translation of Vectors and Slope

To multiply a vector by a scalar, as in Example 5, we don’t actually need to translate it to the origin first; we can just multiply the coordinates of the initial and terminal points of the vector by the scalar k. However, this problem provides useful practice in translating vectors as well as in scalar multiplication.

You may want to clarify that although the same ordered pair can represent many different vectors, it can only represent one vector in standard position. That vector is equivalent to, and in a sense represents, infinitely many other vectors with the same magnitude and direction but different initial points.

Unit Vectors and Components

(Note: \begin{align*}\vec i\end{align*} and \begin{align*}\vec j\end{align*} are typically read as “\begin{align*}i-\end{align*}hat” and “\begin{align*}j-\end{align*}hat,” but often you can call them simply \begin{align*}i\text{''}\end{align*} and \begin{align*}j\text{''}\end{align*} without confusing anyone.)

Resultant as the Sum of Two Components

You may need to slow down to explain the notation here. In the diagram, the blue vector is \begin{align*}r\end{align*} and the green vector is \begin{align*}s\end{align*}, but we are expressing them in a slightly different way to emphasize their role as component vectors. Since vector \begin{align*}r\end{align*} is horizontal, it can be expressed as \begin{align*}|\vec r|\hat i\end{align*}—that is, as the horizontal unit vector \begin{align*}\hat i\end{align*} multiplied by the scalar quantity “the magnitude of \begin{align*}r\end{align*},” yielding a vector with the same magnitude as \begin{align*}r\end{align*} (which makes sense because it is \begin{align*}r\end{align*}!) and in the same direction as the horizontal unit vector \begin{align*}\hat i\end{align*}. By the same logic, vector s can be expressed as \begin{align*}|\vec s| \hat j\end{align*}, because it has the magnitude of \begin{align*}s\end{align*} and is in the direction of \begin{align*}\hat j\end{align*}.

Points to Consider

1) One way to verify answers to an addition or subtraction problem is to resolve each vector into its components and then add or subtract the components separately.

2) Vectors often form oblique triangles when added, so many of the same solving techniques apply.

3) Working with vectors can be more difficult because they are expressed in terms of their relationship to the origin rather than their relationship to each other, so finding things like the angles between them may be harder.

4) Vectors are usually used to solve problems where force is being applied in different directions, as opposed to problems that simply deal with distances between objects.

5) Unit vectors can help us visualize distances on a coordinate grid, but we don’t generally need to use them when we are not looking for such distances.

## Real-World Triangle Problem Solving

Introduction

To clarify situation 1: the climber will need enough rope to reach in a straight line from the top of the wall to the point where his partner will be standing (i.e. the point where he is now), so we need to find that distance to determine the necessary amount of rope.

To clarify situation 2: the axles run lengthwise through the centers of the cylinders and stick out at each end, so the steel cable to hold the cylinders together needs only to run around all three axles, rather than around the outsides of the cylinders. (This will be made clearer by the diagram in the next section.) Also, two loops of cable will need to be wrapped around the axles—one at each end of the cylinders—so once we find the length of one loop of cable, we will need to double it.

Represent Problem Situations as Triangle(s)

In the diagram for situation 3, make sure students realize that the label \begin{align*}\theta\end{align*} applies to the whole span of the angle between the resultant vector and the horizontal line, not just the angle between the resultant vector and the vector next to it.

Make a Problem-Solving Plan

In order to figure out what information they still need to solve each problem, students may first need to think about what tools they might use to solve it. Whether or not the triangle is a right triangle is especially important to consider, as right triangles can be solved with much easier methods.

Choose Among All Available Tools

In situation 1, note that we are rounding our answers to the nearest whole number of feet; this is reasonable since nothing more precise was specified. In situation 2, however, since we started off with values that were rounded to the nearest tenth, it is reasonable to round our answers to the nearest tenth as well.

Points to Consider

1) We could verify our answer in situation 1 using different trig ratios, or trig ratios of different sides, than we used to find the answer. In situation 2, we could use the Law of Cosines to verify the angles that we didn’t find with the Law of Cosines in the first place; in situation 3, we could do that with the laws of both Sines and Cosines. Those would be less reliable methods, though, as we might get mixed up and simply perform the same operations we used to get the answers in the first place, thereby verifying nothing.

2) Checking your answer might not help if you verified the answer with the same methods you used to get the answer—or with those same methods in disguise.

3) The Law of Sines might be unreliable for checking angles because of the Ambiguous Case.

4) Some of the tools used to check them might have been used to solve them; also, applying the same tools in a different order might have worked.

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