# 2.5: Triangles and Vectors

**At Grade**Created by: CK-12

## The Law of Cosines

**In-Text Examples**

1) A common error when working with the Law of Cosines is to forget to subtract, rather than add, the last term \begin{align*}(2ab \cos C)\end{align*}

2) In cases where students know all three side lengths of a triangle and are trying to find one of the angles, they may fail to plug the side lengths into the formula in the correct places. A mnemonic that may help is “the length of the side *opposite* the angle we are trying to find goes on the *opposite* side of the equation from the other two side lengths.”

**Review Questions**

4) Students may forget here that \begin{align*}23.3^\circ\end{align*}

6) Once students have found the correct measure of the angle or side, they may well forget to subtract it from the given measure to tell how far off the given measure is.

10) Part a should read “If the tee is \begin{align*}329\;\mathrm{yards}\ldots\end{align*}

On part b, students may be a little confused about how to set up the new diagram. Basically, it should look like the previous diagram, except with \begin{align*}``98\;\mathrm{yds}\text{''}\end{align*}

12) The instruction “use some right triangle trig” may give some students the idea that the triangle described in the problem is a right triangle. Other students may need to be reminded of the formula for the area of a triangle. Others may overcomplicate the problem by trying to find the measures of more than one angle of the triangle when just one will do.

14) Some students may try to treat this quadrilateral as a parallelogram and find the area based on that formula, instead of finding the areas of the two triangular subsections separately.

15) Students are liable to think that \begin{align*}17^\circ\end{align*}

More rarely, students may think that \begin{align*}17^\circ\end{align*}

16) Since the length of \begin{align*}AB\end{align*}

## Area of a Triangle

**In-Text Examples**

2) It’s very easy, when using Heron’s Formula, to slip up and forget to divide by \begin{align*}2\end{align*}

**Review Questions**

1) In parts \begin{align*}b\end{align*}

Also, they may try to actually find the areas instead of merely stating which formula to use to find them, but that isn’t so bad in this case since it’s what the next question asks them to do anyway. This also applies to problems 3 and 4.

2) Rounding off too early may get some students in trouble here; remind them if necessary not to round off until the very last step so that roundoff errors don’t accumulate.

5) On this problem, students may be tempted to plug in \begin{align*}375\;\mathrm{feet}\end{align*}

6) Students may need to be reminded here that the contractor will have to buy a whole number of bundles.

7) \begin{align*}14955.6\;\mathrm{square\ yards}\end{align*}

9) After plugging the area of the triangle into the \begin{align*}\frac{1}{2} bh\end{align*}

**Additional Problems**

1) Use the identity from problem 10 to prove the Pythagorean Theorem; that is, show that \begin{align*}f^2 = d^2 + e^2\end{align*}

**Answers to Additional Problems**

1) Start with the identity from problem 10:

\begin{align*}d^2 + e^2 + f^2 = 2(ef \cos D + df \cos E + de \cos F)\end{align*}

Now if \begin{align*}F\end{align*}

\begin{align*}d^2 + e^2 + f^2 = 2(ef \cos D + df \cos E)\end{align*}

Drawing the triangle will show us that \begin{align*}\cos D = \frac{e}{f}\end{align*}

\begin{align*}d^2 + e^2 + f^2 = 2 \left (ef \cdot \frac{e} {f} + df \cdot \frac{d}{f} \right )\end{align*}

Canceling out the \begin{align*}fs\end{align*}:

\begin{align*}d^2 + e^2 + f^2 = 2(e^2 + d^2) = 2(d^2 + e^2)\end{align*}

And subtracting \begin{align*}(d^2 + e^2)\end{align*} from both sides:

\begin{align*}f^2 = d^2 + e^2\end{align*}

## The Law of Sines

**Review Questions**

1) Part c is a trick question; it is neither AAS nor ASA but AAA, which we haven’t covered yet.

2) Because of problem 1c, students may assert that the triangles in the chart don’t all have anything in common, or don’t have anything in common except that we know at least two angles. This is technically correct, but the problem is really trying to get at what the ASA and AAS cases have in common, ignoring the AAA case.

Students also shouldn’t be penalized for answering that in both cases we know two angles (or can find three angles) and one side.

3) Again, part c can’t be solved, so don’t let students get stuck on it. (You may want to stress the “if possible” part of the instructions before they get started.)

6) Students may mix up the ASA and SAS cases here, leading them to mix up the Law of Sines and Law of Cosines cases in turn. Also, they may try to actually solve the triangles instead of just stating how they would solve them.

7) The \begin{align*}x\end{align*} we are solving for here is the “other half” of the base of the triangle.

8) After doing all the steps needed to find the various distances, students may forget to subtract the old distance from the new distance, or may forget to calculate the time the extra distance took as their final step.

9) Students may forget to have the driver start at the warehouse rather than at stop \begin{align*}A\end{align*}, or may forget that she needs to get back to the warehouse after stop \begin{align*}C\end{align*}; in other words, they may forget to add in that extra \begin{align*}1.1\;\mathrm{miles}\end{align*} of distance either or both times.

Also, they may label the angles wrong, for example labeling the \begin{align*}103^\circ\end{align*} angle at stop \begin{align*}B\end{align*} as the exterior angle rather than the interior angle of the triangle; this is an easy mistake to make here, since they may unconsciously be thinking of the angles the two other streets make with First Street as both being in “standard position” because First Street is horizontal.

Finally, they may forget to add in the \begin{align*}2\;\mathrm{minutes}\end{align*} for each package when calculating the time, or may add \begin{align*}2\;\mathrm{minutes}\end{align*} just once instead of three times. Or, they may get the time constraints backwards, thinking of the driver as leaving at 10:00 and trying to calculate when she will get back, instead of calculating when she must leave to get back by 10:00.

10) There is more than one way to set up this problem, depending on how one interprets the descriptions given. Students therefore may not get the answer given, but should not be penalized if they can show that they have set up the problem in a way that seems reasonable.

## The Ambiguous Case

**Review Questions**

3) Students may approach this problem in reverse, by rewriting the given equation as \begin{align*}\frac{a} {c} - \frac{c} {c} = \frac{\sin A} {\sin C} - \frac{\sin C}{\sin C}\end{align*} and going from there, ending up with the Law of Sines as their final step. This is an equally valid solution.

4) Students may have a hard time coming up with an appropriate set of sides and angle if they try picking the sides first; they will have much better luck if they think of starting with an angle \begin{align*}A\end{align*} and a side \begin{align*}b\end{align*} and then finding a side \begin{align*}a\end{align*} that is less than \begin{align*}b\end{align*} and greater than \begin{align*}b \sin A\end{align*}. Even then, though, this will only work if they pick an acute \begin{align*}A\end{align*}, although it may not be immediately obvious why. (Drawing a picture may make it clearer: if \begin{align*}A\end{align*} is the biggest angle, then \begin{align*}a\end{align*} must be the biggest side and certainly can’t be smaller than \begin{align*}b\end{align*}.

5) Students may think there is only one value of \begin{align*}A\end{align*} in each case, rather than a whole range of values.

6) Solving this problem without using the Law of Sines at all is actually possible, and students should probably not be penalized for doing so.

7) Students may get stuck if they try to find all the sides and angles in the order specified; they should be encouraged to find them in whatever order they can. Also, make sure they realize that it’s \begin{align*}\angle ABC\end{align*} that measures \begin{align*}109.6^\circ\end{align*}, rather than any of the other angles whose vertices are at \begin{align*}B\end{align*}.

8) Some students may jump to the conclusion that the triangle shown is a right triangle, simply because it looks like one.

10) The last question can’t actually be answered using the information given.

**Additional Problems**

1) In the figure below, \begin{align*}AB = 11.5, BE =10.3, EC = 7.8, CD = 8.1, \angle AEB = 50.1^\circ,\end{align*} and \begin{align*}\angle CED = 42.7^\circ.\end{align*}

Find the following, to the nearest tenth of a unit:

a) \begin{align*}\angle BEC\end{align*}

b) \begin{align*}BC\end{align*}

c) \begin{align*}\angle EBC\end{align*}

d) \begin{align*}\angle ECB\end{align*}

e) \begin{align*}\angle BAE\end{align*}

f) \begin{align*}\angle ABE\end{align*}

g) \begin{align*}AE\end{align*}

h) \begin{align*}\angle EDC\end{align*}

i) \begin{align*}\angle ECD\end{align*}

j) \begin{align*}ED\end{align*}

**Answers to Additional Problems**

1) a) \begin{align*}87.2^\circ\end{align*} (because \begin{align*}\angle AED\end{align*} is a straight angle)

b) \begin{align*}12.6\end{align*} (by Law of Cosines)

c) \begin{align*}38.2^\circ\end{align*} (by Law of Sines)

d) \begin{align*}54.7^\circ\end{align*} (by Law of Sines)

e) \begin{align*}43.4^\circ\end{align*} (by Law of Sines)

f) \begin{align*}86.5^\circ\end{align*} (by Triangle Sum Theorem)

g) \begin{align*}15.0\end{align*} (by Law of Sines)

h) \begin{align*}40.8^\circ\end{align*} (by Law of Sines)

i) \begin{align*}96.5^\circ\end{align*} (by Triangle Sum Theorem)

j) \begin{align*}11.9\end{align*} (by Law of Sines)

## General Solutions of Triangles

**Review Questions**

3) On part d, students may think that, as on part e, we are still missing side \begin{align*}c\end{align*} and angle \begin{align*}C\end{align*}. However, the fact that there is no solution possible to this triangle actually means that there is no such triangle, so technically we are not “missing” that last side and angle measure because the side and angle themselves do not exist.

5) Students may need to be reminded that a rhombus has all four sides the same and that both pairs of opposite angles are congruent. They may also need to be reminded that they can use Heron’s formula to find the area of half of the rhombus, and they may forget to double that area to find that of the whole rhombus.

6) The likeliest error here, once students realize they need to divide the pentagon into triangles to get anywhere with it at all, is for them to divide it into triangles they can’t solve. This will happen if they connect the vertices whose angles are already known, dividing them up into unknown angles. Instead, they need to connect the three vertices whose angles aren’t given; then they will have triangles they can solve.

7) There isn’t enough information given to solve the triangle; you may need to supply a couple of angle or side measures from the answer key so that students can find the rest.

**Additional Problems**

1) Find the area of the quadrilateral below, to the nearest hundredth.

2) Find the missing angle measures in the above quadrilateral, to the nearest tenth.

**Answers to Additional Problems**

1) First, divide the quadrilateral into two triangles by drawing diagonal \begin{align*}AC\end{align*}. Then the area of each triangle can be found with \begin{align*}K = \frac{1}{2} bc \sin A\end{align*}.

The area of \begin{align*}\triangle ABC\end{align*} is \begin{align*}22.57\end{align*}, and the area of \begin{align*}\triangle ADC\end{align*} is \begin{align*}33.85\end{align*}. The total area of the quadrilateral is therefore \begin{align*}56.42\end{align*}.

2) The fact that \begin{align*}ABCD\end{align*} is symmetrical is what makes this problem possible to solve. Divide \begin{align*}ABCD\end{align*} into two triangles by drawing diagonal \begin{align*}BD\end{align*}; these two triangles are congruent by SSS, and so each of them has area \begin{align*}28.21\end{align*} (half of the area of \begin{align*}ABCD\end{align*}). The area formula \begin{align*}K = \frac{1}{2} bc \sin A\end{align*} can now be used to find the missing angle in each triangle.

Both missing angles, \begin{align*}\angle A\end{align*} and \begin{align*}\angle C\end{align*}, measure \begin{align*}88.5^\circ\end{align*}.

## Vectors

**In-Text Examples**

1) Stress that the distance formula is just a version of the Pythagorean Theorem: the magnitude of the vector can be thought of as \begin{align*}\sqrt{(\mathrm{difference\ between\ x-coordinates})^2 + (\mathrm{difference\ between\ y-coordinates)}^2}\end{align*}.

**Review Questions**

1) The answers given in the text include the vector directions with respect to \begin{align*}\vec{m}\end{align*}—that is, the direction of each resultant vector is expressed as the angle the vector makes with \begin{align*}\vec{m}\end{align*}—so students will get the same answers as those in the text if they too express their resultant vectors in terms of \begin{align*}\vec{m}\end{align*}. If you don’t specifically tell them to do this, though, they may express them in terms of \begin{align*}\vec n\end{align*} instead, in which case the angles they give as their answers should be the complements of the angles given in the book. (Either way is correct in the absence of more specific instructions.)

2) Students can use either the triangle or parallelogram method here. Since they will need to use a ruler and protractor, make sure their copy of the text is the right size or they will get the wrong answers. (It may be worth double-checking by measuring the lengths of the vectors \begin{align*}\vec a\end{align*} through \begin{align*}\vec d\end{align*} as they appear on the page.)

Also, note that the diagrams on the first two lines of the chart are not to scale; students will have to re-draw the vectors to get the correct magnitude and direction.

And finally, they will be measuring angles from the horizontal when there isn’t a horizontal line drawn in the text, so they will have to estimate them as best they can; don’t be too strict about their getting the angles exactly right.

3) This is actually only true if \begin{align*}\vec a\end{align*} and \begin{align*}\vec b\end{align*} have the same or opposite direction.

4) The direction angle, \begin{align*}4.6^\circ\end{align*} NW, can also be expressed as an angle of \begin{align*}94.6^\circ\end{align*} in standard position.

8-10) Students will have to draw some fairly large vectors to solve these problems, and they will still need to scale them down in order for them to fit on one page. As a result, they may not be able to get their answers accurate to the nearest tenth or even the nearest whole unit. (They won’t have this problem if they use the Law of Cosines and the Law of Sines instead of drawing the vectors, but since that technique hasn’t been covered yet, it probably won’t occur to them to use it and they certainly shouldn’t be expected to.)

Also, on problems 8 and 10, note that the angle between the two vectors in each case is the angle between them when they are placed tail-to-tail. This makes the parallelogram method fairly easy to apply, but students who use the triangle method instead may think that the angle given is the angle between the vectors when they are placed tip-to-tail, and will get the wrong answer as a result.

## Component Vectors

**In-Text Examples**

5) Simply multiplying the coordinates by \begin{align*}2.5\end{align*} without translating first yields a directed segment from \begin{align*}(10, 17.5)\end{align*} to \begin{align*}(30, 27.5)\end{align*}. This answer isn’t wrong, as it has the same magnitude and direction as the one from \begin{align*}(4, 7)\end{align*} to \begin{align*}(24, 17)\end{align*}, but you may want to make sure students understand that the answers are equivalent.

**Review Questions**

2) Students may try to solve these problems by translating the vectors to the origin and reading off the coordinates of the new terminal point, or they may try to solve them by finding the difference between the \begin{align*}x-\end{align*}coordinates of the two points and the difference between the \begin{align*}y-\end{align*}coordinates of the two points. It may be worth pointing out that these two methods are in fact mathematically equivalent; both involve performing essentially the same operations and both yield the same answer.

8) This question may confuse students a bit; some of them may try to solve it as if they were given two components and asked for the resultant vector, while others may try to solve it as if they were given the resultant and one component and asked for the other component. Since the text isn’t entirely clear on which one is actually being asked for, it’s best to accept any answer that a student can justify.

9) This question is similarly ambiguous; as a result, students may get \begin{align*}11.5^\circ\end{align*} rather than \begin{align*}11.3^\circ\end{align*} as their answer, and this should be accepted.

10) You may need to clarify the notation here: \begin{align*}AB\end{align*} means simply the directed line segment from point \begin{align*}A\end{align*} to point \begin{align*}B\end{align*}.

**Additional Problems**

1) Find the single ordered pair that represents \begin{align*}\vec a\end{align*} in each equation if you are given \begin{align*}\vec b = (1, 2)\end{align*} to \begin{align*}(3, 8)\end{align*} and \begin{align*}\vec c = (3, 3)\end{align*} to \begin{align*}(5, 2)\end{align*}.

a) \begin{align*}\vec a = \frac{1}{2} \vec b\end{align*}

b) \begin{align*}\vec a = -5 \vec c\end{align*}

c) \begin{align*}\vec a = \vec b - \vec c\end{align*}

d) \begin{align*}\vec a = 3 \vec c + 2 \vec b\end{align*}

**Answers to Additional Problems**

1) (By translating \begin{align*}\vec b \end{align*} and \begin{align*}\vec c\end{align*} to the origin, you can represent each of them as a single ordered pair, which makes it easier to represent \begin{align*}\vec a\end{align*} as a single ordered pair.)

a) \begin{align*}(1, 3)\end{align*}

b) \begin{align*}(-10, 5)\end{align*}

c) \begin{align*}(0, 7)\end{align*}

d) \begin{align*}(10, 9)\end{align*}

## Real-World Triangle Problem Solving

**Review Questions**

2) To solve this problem, we need to assume that the height of the canyon wall on the hiker’s side is the same height as on the opposite side. (Most students will assume this anyway, but those who don’t should be told to.)

5) Some few non-pool-playing students may need to be told which ball is which in the diagram, and may need to be told that the pockets are at the corners of the table.

7) Some students may not immediately see that they need to find the area of the triangle between the three docks. Others may find the area and forget to multiply by \begin{align*}5.2 \times 10^{13}\end{align*} to find the number of bacteria.

8) Students may jump to the conclusion that \begin{align*}37^\circ\end{align*} is the measure of the angle from tower \begin{align*}B\end{align*} to tower \begin{align*}A\end{align*} to the fire (it’s actually the angle’s complement.) They may also jump to the conclusion that the angle from tower \begin{align*}A\end{align*} to tower \begin{align*}B\end{align*} to the fire is a right angle.

10) Students may think the angle between the first part of the trip and the second part of the trip is \begin{align*}34^\circ\end{align*}. Instead, they need to think in terms of vectors; they should draw the two halves of the trip as individual vectors with their separate headings, and then find their resultant.

**Additional Problems**

1) a) A ship approaches close enough to shore to spot a famous lighthouse which is known to be \begin{align*}180\;\mathrm{feet}\end{align*} tall, and which stands at the top of a \begin{align*}2200-\end{align*}foot cliff. The captain, looking from a point \begin{align*}15\;\mathrm{feet}\end{align*} above the water, observes that the angle of elevation to the top of the lighthouse is \begin{align*}4.74^\circ\end{align*}. How far is the ship from the lighthouse, to the nearest foot? To the nearest tenth of a mile? \begin{align*}(5280\;\mathrm{feet} = 1\;\mathrm{mile}.)\end{align*}

b) The lighthouse keeper notices that the ship appears to be headed toward a dangerous reef. The reef is known to be \begin{align*}2.8\;\mathrm{miles}\end{align*} from the lighthouse, and the lighthouse at this moment makes an angle of \begin{align*}30.8^\circ\end{align*} with the ship and the reef. How close is the ship to the reef?

**Answers to Additional Problems**

1) a)

\begin{align*}\tan 4.74^\circ = \frac{2370}{x} \ \ x = \frac{2370}{\tan 4.74^\circ} \ \ x = 28,522\end{align*}

The ship is \begin{align*}28,522\;\mathrm{feet}\end{align*}, or \begin{align*}4.9\;\mathrm{miles}\end{align*}, from the lighthouse.

b)

Using the Law of Cosines:

\begin{align*}x^2 & = (4.9)^2 + (2.8)^2 - (2 \times 4.9 \times 2.8 \cos (30.8^\circ))\\ \therefore x & \approx 2.9\end{align*}

The ship is \begin{align*}2.9\;\mathrm{miles}\end{align*} from the reef.

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