My Favorite One-Liners: Part 9

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Today, I’d like to discuss a common mistake students make in trigonometry… as well as the one-liner that I use to (hopefully) help students not make this mistake in the future.

Question. Find all solutions (rounded to the nearest tenth of a degree) of \sin x = 0.8.

Erroneous Solution. Plugging into a calculator, we find that x \approx 53.1^o.

arcsine1

The student correctly found the unique angle x between -90^o and 90^o so that \sin x = 0.8. That’s the definition of the arcsine function. However, there are plenty of other angles whose sine is equal to 0.7. This can happen in two ways.

First, if $\sin x > 0$, then the angle x could be in either the first quadrant or the second quadrant (thanks to the mnemonic All Scholars Take Calculus). So x could be (accurate to one decimal place) equal to either 53.1^o or else 180^o - 53.1^o = 126.9^o. Students can visualize this by drawing a picture, talking through each step of its construction (first black, then red, then brown, then green, then blue).arcsin45

However, most students don’t really believe that there’s a second angle that works until they see the results of a calculator.

TIarcsin45

Second, any angle that’s coterminal with either of these two angles also works. This can be drawn into the above picture and, as before, confirmed with a calculator.

So the complete answer (again, approximate to one decimal place) should be 53.1^{\circ} + 360n^o and 126.9 + 360n^{\circ}, where n is an integer. Since integers can be negative, there’s no need to write \pm in the solution.

Therefore, the student who simply answers 53.1^o has missed infinitely many solutions. The student has missed every nontrivial angle that’s coterminal with 53.1^o and also every angle in the second quadrant that also works.

green line

Here’s my one-liner — which never fails to get an embarrassed laugh — that hopefully helps students remember that merely using the arcsine function is not enough for solving problems such as this one.

You’ve forgotten infinitely many solutions. So how many points should I take off?

For further reading, here’s my series on inverse functions.

Engaging students: Inverse trigonometric functions

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Joe Wood. His topic, from Precalculus: inverse trigonometric functions.

green line

What are the contributions of various cultures to this topic?

Trig functions have a very long history spanning many countries and cultures. Greek astronomers such as Aristarchus, Claudius, and Ptolemy first used trigonometry; however, according to the University of Connecticut, these Greek astronomers were primarily concerned with “the length of the chord of a circle as a function of the circular arc joining its endpoints.” Many of these astronomers, Ptolemy especially, were concerned with planetary and celestial body’s rotations, so this made sense.

While the Greeks first studied trigonometric concepts, it was the Indian people who really studied sine and cosine functions with the angle as a variable. The information was then brought to the Arabic and Persian cultures. One significant figure, a Persian by the name Abu Rayhan Biruni, used trig to accurately estimate the circumference of Earth and its radius before the end of the 11th century.

Fast-forward about 700 years, a Swiss mathematician, Daniel Bernoulli, used the “A.sin” notation to represent the inverse of sine. Shortly after, another Swiss mathematician used “A t” to represent the inverse of tangent. That man was none other than Leonhard Euler.  It was not until 1813 that the notation sin-1 and tan-1 were introduced by Sir John Fredrick William Herschel, an English mathematician.

As we can see, the development of inverse trigonometric functions took quite the cultural rollercoaster ride before stopping some place we see being familiar. It took many cultures, and even more years to develop this sophisticated branch of mathematics.

 

green line

How could you as a teacher create an activity or project that involves your topic?

Last Semester I taught a lesson on the trigonometric identities. I found this cool cut and paste activity for the students that allowed them to warm up to the trig identities by not having to do the process themselves, but still having to see every step of converting one trig function into another with the identities. Below, you will find the activity, then the instructions, and finally how to modify the activity to fit inverse trig identities specifically.

inversetrig

Directions:
1.) Begin by cutting out all the pieces.
2.) Students will take any of the four puzzle pieces with the black squiggly line.
3.) Find an equivalent puzzle piece by using some trig identity.
4.) Repeat step 3 until there are no more equivalent pieces.
5.) Grab the next puzzle pieces with the black squiggly line.
6.) Repeat steps 3-5 until all puzzle pieces have been used.
Ex.) Begin with cscx-sinx. Lay next to that piece, the piece that reads =1/sinx – sinx, then the piece that reads =1/sinx – sin2x/sinx. Contiue the trend until you reach =cotx * cosx. Then move to the next squiggly lined piece.

Modify:
This game can be modified using inverse trig functions. Start with pieces such as sin-1(sin(300)) in squiggles. Have a piece showing sin-1(sin(300)) with a line through the sines. Then a piece that just shows 300. Next a piece in a squiggly line that is sin-1(sqrt(2)/2) that connects to a piece of 450, but make them write why this works.

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How does this topic extend what your students should have learned in previous courses?

Obviously, by this time students should know what trigonometric functions are and how to use this. Students should also know from previous classes what inverse functions are. Studying inverse trig functions then is a continuation of these topics. As I teacher I would begin relating inverse trig functions by refreshing the students on what inverse functions are. The class would then move into the concept that the trig expression of an angle returns a ratio of two sides of a triangle. We would slowly move into what happens then if you know the sides of a triangle but need the angle. From there we would discuss trigonometric expressions using the angles as variables. Finally, we would make the connection that that is a function, and on the proper interval should have an inverse function. That is when the extension into the new topic of inverse trigonometric functions would seriously begin.

 

Two ways of doing an integral: Index

I’m using the Twelve Days of Christmas (and perhaps a few extra days besides) to do something that I should have done a long time ago: collect past series of posts into a single, easy-to-reference post. The following posts formed my series on different ways of computing the integral \displaystyle \int \frac{dx}{\sqrt{4x-x^2}}.

Part 1: The two “different” answers.

Part 2: Explaining why the two “different” answers are really equivalent.

 

 

 

Inverse Functions: Arccosine and SSS (Part 21)

Arccosine has an important advantage over arcsine when solving for the parts of a triangle: there is no possibility ambiguity about the angle.

Solve \triangle ABC if a = 16, b = 20, and c = 25.

When solving for the three angles, it’s best to start with the biggest angle (that is, the angle opposite the biggest side). To see why, let’s see what happens if we first use the Law of Cosines to solve for one of the two smaller angles, say \alpha:

a^2 = b^2 + c^2 - 2 b c \cos \alpha

256 = 400 + 625 - 1000 \cos \alpha

-769 = -1000 \cos \alpha

0.769 = \cos \alpha

\alpha \approx 39.746^\circ

So far, so good. Now let’s try using the Law of Sines to solve for \gamma:

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}

\displaystyle \frac{\sin 39.746^\circ}{16} \approx \displaystyle \frac{\sin \gamma}{25}

0.99883 \approx \sin \gamma

Uh oh… there are two possible solutions for \gamma since, hypothetically, \gamma could be in either the first or second quadrant! So we have no way of knowing, using only the Law of Sines, whether \gamma \approx 87.223^\circ or if \gamma \approx 180^\circ - 87.223^\circ = 92.777^\circ.

green lineFor this reason, it would have been far better to solve for the biggest angle first. For the present example, the biggest answer is \gamma since that’s the angle opposite the longest side.

c^2 = a^2 + b^2 - 2 a b \cos \gamma

625 = 256 + 400 - 640 \cos \gamma

-31 =-640 \cos \gamma

0.0484375 = \cos \gamma

Using a calculator, we find that \gamma \approx 87.223^\circ.

We now use the Law of Sines to solve for either \alpha or \beta (pretending that we didn’t do the work above). Let’s solve for \alpha:

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}

\displaystyle \frac{\sin \alpha}{16} \approx \displaystyle \frac{\sin 87.223}{25}

\sin \alpha \approx 0.63949

This equation also has two solutions in the interval [0^\circ, 180^\circ], namely, \alpha \approx 39.736^\circ and \alpha \approx 180^\circ - 39.736^\circ = 140.264^\circ. However, we know full well that the answer can’t be larger than \gamma since that’s already known to be the largest angle. So there’s no need to overthink the matter — the answer from blindly using arcsine on a calculator is going to be the answer for \alpha.

Naturally, the easiest way of finding \beta is by computing 180^\circ - \alpha - \gamma.

Inverse Functions: Arccosine and SSS (Part 19)

Arccosine has an important advantage over arcsine when solving for the parts of a triangle: there is no possibility ambiguity about the angle.

Solve \triangle ABC if latex a = 16$, b = 20, and c = 25.

To solve for, say, the angle \gamma, we employ the Law of Cosines:

 

c^2 = a^2 + b^2 - 2 a b \cos \gamma

625 = 256 + 400 - 640 \cos \gamma

-31 =-640 \cos \gamma

0.0484375 = \cos \gamma

Using a calculator, we find that \gamma \approx 87.2^\circ. And the good news is that there is no need to overthink this… this is guaranteed to be the angle since the range of y = \cos^{-1} x is [0,\pi], or [0^\circ, 180^\circ] in degrees. So the equation

\cos x = \hbox{something}

is guaranteed to have a unique solution between 0^\circ and 180^\circ. (But there are infinitely many solutions on \mathbb{R}. And since an angle in a triangle must lie between 0^\circ and 180^\circ, the practical upshot is that just plugging into a calculator blindly is perfectly OK for this problem. This is in stark contrast to the Law of Sines, for which some attention must be paid for solutions in the interval [0^\circ,90^\circ] and also the interval [90^\circ, 180^\circ].

From this point forward, the Law of Cosines could be employed again to find either \alpha or \beta. Indeed, this would be my preference since the sides a, b, and c are exactly. However, my experience is that students prefer the simplicity of the Law of Sines to solve for one of these angles, using the now known pair of c (exactly known) and \gamma (approximately known with a calculator).

Inverse functions: Arcsine and SSA (Part 17)

In the last few posts, we studied the SSA case of solving for a triangle, when two sides and an non-included angle are given. (Some mathematics instructors happily prefer the angle-side-side acronym to bluntly describe the complications that arise from this possibly ambiguous case. I personally prefer not to use this acronym.)

A note on notation: when solving for the parts of \triangle ABC, a will be the length of the side opposite \angle A, b will be the length of the side opposite \angle B, and c will be the length of the side opposite angle C. Also \alpha will be the measure of \angle A, \beta will be measure of \angle B, and \gamma will be the measure of \angle C. Modern textbooks tend not to use \alpha, \beta, and \gamma for these kinds of problems, for which I have only one response:https://meangreenmath.files.wordpress.com/2014/10/philistines.png

philistines

Suppose that a, c, and the nonincluded angle \alpha are given, and we are supposed to solve for b, \beta, and \gamma. As we’ve seen in this series, there are four distinct cases — and handling these cases requires accurately solving equation like \sin \gamma = \hbox{something} on the interval [0^\circ, 180^\circ].

Case 1. b < c \sin \alpha. In this case, there are no solutions. When the Law of Sines is employed and we reach the step

\sin \gamma = \hbox{something}

the \hbox{something} is greater than 1, which is impossible.

SSA1

Case 2. b = c \sin \alpha. This rarely arises in practice (except by careful writers of textbooks). In this case, there is exactly one solution. When the Law of Sines is employed, we obtain

\sin \gamma = 1

We conclude that \gamma = 90^\circ, so that \triangle ABC is a right triangle.

SSA2

Case 3. c \sin \alpha < b < c. This is the ambiguous case that yields two solutions. The Law of Sines yields

\sin \gamma = \hbox{something}

so that there are two possible choices for \gamma, \hbox{some angle} and 180^\circ - \hbox{some angle}.

SSA4

Case 4. b > c. This yields one solution. Similar to Case 3, the Law of Sines yields

\sin \gamma = \hbox{something}

so that there are two possible choices for \gamma, \hbox{some angle} and 180^\circ - \hbox{some angle}. However, when the second larger value of \gamma is attempted, we end up with a negative angle for \beta, which is impossible (unlike Case 3).

 

SSA3Many mathematics students prefer to memorize rules like those listed above. However, I try to encourage my students not to blindly use rules when solving the SSA case, as it’s just too easy to make a mistake in identifying the proper case. Instead, I encourage them to use the Law of Sines and to remember that the equation

\sin \gamma = t

has two solutions in [0^\circ, 180^\circ] as long as $0 < t < 1$:

\gamma = \sin^{-1} t \qquad \hbox{and} \qquad \gamma = 180^\circ - \sin^{-1} t

If they can remember this fact, then students can just follow their noses when applying the Law of Sines, identifying impossible and ambiguous cases when the occasions arise.

 

Inverse functions: Arcsine and SSA (Part 16)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like \sin \theta = 0.8. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of \triangle ABC, a will be the length of the side opposite \angle A, b will be the length of the side opposite \angle B, and c will be the length of the side opposite angle C. Also \alpha will be the measure of \angle A, \beta will be measure of \angle B, and \gamma will be the measure of \angle C. Modern textbooks tend not to use \alpha, \beta, and \gamma for these kinds of problems, for which I have only one response:

philistines

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has two different solutions:

Solve \triangle ABC if a = 8, c = 10, and \alpha = 30^\circ.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

badSSA4

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

SSA4This time, the red circle intersects the dashed black line at two different points. So there will be two different solutions for this case. In other words, the phrasing of the question is somewhat deceptive. Usually when the question asks, “Solve the triangle…”, it’s presumed that there is only one solution. In this case, however, there are two different solutions.

These two different solutions appear when using the Law of Sines:

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}

\displaystyle \frac{\sin 30^\circ}{8} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle \frac{1/2}{8} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle \frac{5}{8} = \sin \gamma

At this point, the natural inclination of a student is to pop out the calculator and find \sin^{-1} \frac{1}{3}.

SSAcalc4

This is incorrect logic that, as discussed extensively in earlier in this series of posts, there are two angles between 0^\circ and 180^\circ with a sine of 5/8:

\sin^{-1} \frac{5}{8} \qquad \hbox{and} \qquad \pi - \sin^{-1} \frac{5}{8},

or, in degrees,

\gamma \approx 38.68^\circ \qquad \hbox{and} \qquad \gamma \approx 141.32^\circ

So we have two different cases to check. Unlike the previous posts in this series, it’s really, really important that we list both of these cases.

Case 1: \gamma \approx 38.68^\circ. We begin by solving for \beta:

\beta = 180^\circ - \alpha - \gamma \approx 111.32^\circ

Then we can use the Law of Sines to find b. In this case, it’s best to use the pair \alpha - a instead of \gamma - c since the values of \alpha and a are both known exactly.

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \beta}{b}

\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin 111.32^\circ}{b}

b = \displaystyle \frac{8 \sin 111.32^\circ}{\sin 30^\circ}

b \approx 14.9

This triangle with \gamma \approx 38.68^\circ, \beta \approx 111.32^\circ, and b \approx 14.9 corresponds to the bigger of the two triangles in the above picture, or the rightmost of the two places where the dotted circle intersects the black dotted line.

Case 2: \gamma \approx 141.32^\circ. We again begin by solving for \beta:

\beta = 180^\circ - \alpha - \gamma \approx 8.68^\circ

Unlike yesterday’s example, this is possible. So we have to continue the calculation to find b:

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \beta}{b}

\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin 8.68^\circ}{b}

b = \displaystyle \frac{8 \sin 8.68^\circ}{\sin 30^\circ}

b \approx 2.4

This second triangle with \gamma \approx 141.32^\circ, \beta \approx 8.68^\circ, and b \approx 2.4 corresponds to the thinner of the two triangles in the above picture, or the leftmost of the two places where the dotted circle intersects the black dotted line.

Inverse functions: Arcsine and SSA (Part 15)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like \sin \theta = 0.8. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of \triangle ABC, a will be the length of the side opposite \angle A, b will be the length of the side opposite \angle B, and c will be the length of the side opposite angle C. Also \alpha will be the measure of \angle A, \beta will be measure of \angle B, and \gamma will be the measure of \angle C. Modern textbooks tend not to use \alpha, \beta, and \gamma for these kinds of problems, for which I have only one response:

philistines

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has exactly one solution:

Solve \triangle ABC if a = 15, c = 10, and \alpha = 30^\circ.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

badSSA3

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

SSA3

Notice that the red circle intersects the dashed black line at exactly one point. Therefore, we know that there will be exactly one solution for this case. We also note that the circle would have intersected the black dashed line had the dashed line been extended to the left. This will become algebraically clear in the solution below.

Of course, students should not be expected to make a picture this accurately when doing homework. Fortunately, this impossibility naturally falls out of the equation when using the Law of Sines:

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}

\displaystyle \frac{\sin 30^\circ}{15} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle \frac{1/2}{15} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle \frac{1}{3} = \sin \gamma

At this point, the natural inclination of a student is to pop out the calculator and find \sin^{-1} \frac{1}{3}.

SSAcalc1

This is incorrect logic that, as discussed extensively in yesterday’s post, there are two angles between 0^\circ and 180^\circ with a sine of 1/3:

\sin^{-1} \frac{1}{3} \qquad \hbox{and} \qquad \pi - \sin^{-1} \frac{1}{3},

or, in degrees,

\gamma \approx 19.47^\circ \qquad \hbox{and} \qquad \gamma \approx 160.53^\circ

So we have two different cases to check.

Case 1: \gamma \approx 19.47^\circ. We begin by solving for \beta:

\beta = 180^\circ - \alpha - \gamma \approx 130.53^\circ

Then we can use the Law of Sines (or, in this case, the Pythagorean Theorem), to find b. In this case, it’s best to use the pair \alpha - a instead of \gamma - c since the values of \alpha and a are both known exactly.

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \beta}{b}

\displaystyle \frac{\sin 30^\circ}{15} = \displaystyle \frac{\sin 130.53^\circ}{b}

b = \displaystyle \frac{15 \sin 130.53^\circ}{\sin 30^\circ}

b \approx 22.8

Case 2: \gamma \approx 130.53^\circ. We again begin by solving for \beta:

\beta = 180^\circ - \alpha - \gamma \approx -10.53^\circ

Oops. That’s clearly impossible. So there is only one possible triangle, and the missing pieces are \gamma \approx 19.47^\circ, \beta \approx 130.53^\circ, and b \approx 22.8. Judging from the above (correctly drawn) picture, these numbers certainly look plausible.

It turns out that Case 2 will always fail in SSA will always fail as long as the side opposite the given angle is longer than the other given side (in this case, a > c). However, I prefer that my students not memorize this rule. Instead, I’d prefer that they list the two possible values of \gamma and then run through the logical consequences, stopping when an impossibility is reached. As we’ll see in tomorrow’s post, it’s perfectly possible for Case 2 to produce a second valid solution with the proper choice for the length of the side opposite the given angle.

Inverse functions: Arcsine and SSA (Part 14)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like \sin \theta = 0.8. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of \triangle ABC, a will be the length of the side opposite \angle A, b will be the length of the side opposite \angle B, and c will be the length of the side opposite angle C. Also \alpha will be the measure of \angle A, \beta will be measure of \angle B, and \gamma will be the measure of \angle C. Modern textbooks tend not to use \alpha, \beta, and \gamma for these kinds of problems, for which I have only one response:

philistines

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has exactly one solution:

Solve \triangle ABC if a = 15, c = 10, and \alpha = 30^\circ.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

badSSA3

Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

SSA3

Notice that the red circle intersects the dashed black line at exactly one point. Therefore, we know that there will be exactly one solution for this case. We also note that the circle would have intersected the black dashed line had the dashed line been extended to the left. This will become algebraically clear in the solution below.

Of course, students should not be expected to make a picture this accurately when doing homework. Fortunately, this impossibility naturally falls out of the equation when using the Law of Sines:

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}

\displaystyle \frac{\sin 30^\circ}{15} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle \frac{1/2}{15} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle \frac{1}{3} = \sin \gamma

At this point, the natural inclination of a student is to pop out the calculator and find \sin^{-1} \frac{1}{3}.

SSAcalc1

This is incorrect logic that, as we’ll see tomorrow, nevertheless leads to the correct conclusion. This is incorrect logic because there are two angles between 0^\circ and 180^\circ with a sine of 1/3. There is one solution in the first quadrant (the unique answer specified by arcsine), and there is another answer in the second quadrant — which is between 90^\circ and 180^\circ and hence not a permissible value of arcsine. Let me demonstrate this in three different ways.

First, let’s look at the graph of y = \sin x (where, for convenience, the units of the x-axis are in degrees). This graph intersects the line y = \frac{1}{3} in two different places between 0^\circ and 180^\circ. This does not violate the way that arcsine was defined — arcsine was defined using the restricted domain [-\pi/2,\pi/2], or [-90^\circ, 90^\circ] in degrees.

sinewaveSSA

Second, let’s look at drawing angles in standard position. The angle in the second quadrant is clearly the reflection of the angle in the first quadrant through the y-axis.

standardSSA

Third, let’s use a trigonometric identity to calculate \sin \left( \pi - \sin^{-1} \displaystyle \frac{1}{3} \right):

\sin \left( \pi - \sin^{-1} \displaystyle \frac{1}{3} \right) = \sin \pi \cos \left( \sin^{-1} \displaystyle \frac{1}{3} \right) - \cos \pi \sin \left( \sin^{-1} \displaystyle \frac{1}{3} \right)

=0 \cdot \cos \left( \sin^{-1} \displaystyle \frac{1}{3} \right) + 1 \cdot \sin \left( \sin^{-1} \displaystyle \frac{1}{3} \right)

= \displaystyle \frac{1}{3}

Fourth, and perhaps most convincingly for modern students (to my great frustration), let’s use a calculator:

SSAcalc3

 All this to say, blinding computing \sin^{-1} \frac{1}{3} uses incorrect logic when solving this problem.

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Tomorrow, we’ll examine what happens when we try to solve the triangle using these two different solutions for \gamma.

Inverse functions: Arcsine and SSA (Part 13)

We’ve seen in this series that blinding using the arcsine function on a calculator is insufficient for finding all solutions of an equation like \sin \theta = 0.8. In today’s post, I discuss one of the first places that this becomes practically important: solving the ambiguous case of solving a triangle given two sides and an nonincluded angle.

A note on notation: when solving for the parts of \triangle ABC, a will be the length of the side opposite \angle A, b will be the length of the side opposite \angle B, and c will be the length of the side opposite angle C. Also \alpha will be the measure of \angle A, \beta will be measure of \angle B, and \gamma will be the measure of \angle C. Modern textbooks tend not to use \alpha, \beta, and \gamma for these kinds of problems, for which I have only one response:

philistines

Why does an SSA triangle produce an ambiguous case (unlike the SAS, SSS, or ASA cases)? Here’s a possible problem that has exactly one solution:

Solve \triangle ABC if a = 5, c = 10, and \alpha = 30^\circ.

A student new to the Law of Sines might naively start solving the problem by drawing something like this:

badSSA2Of course, that’s an inaccurate picture that isn’t drawn to scale. A more accurate picture would look like this:

SSA2

Notice that the red circle intersects the dashed black line at exactly one point. Therefore, we know that there will be exactly one solution for this case.

Of course, the reason that the dashed circle and line intersect at exactly one point is because a = c \sin \alpha, so that the triangle is a right triangle.

Of course, students should not be expected to make a picture this accurately when doing homework. Fortunately, this impossibility naturally falls out of the equation when using the Law of Sines:

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \gamma}{c}

\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle \frac{1/2}{5} = \displaystyle \frac{\sin \gamma}{10}

\displaystyle 1 = \sin \gamma

90^\circ = \gamma

The jump to the last step is only possible because there’s exactly one angle between 0^\circ and 90^\circ whose sine is equal to 1. In the next couple posts in this series, we’ll see what happens when we get a step where 0 < \sin \gamma < 1.

Anyway, for the problem at hand, from this point forward it’s easy to solve for the remaining pieces. We begin by finding \beta:

\beta = 180^\circ - \alpha - \gamma = 60^\circ

Then we can use the Law of Sines (or, in this case, the Pythagorean Theorem), to find b:

\displaystyle \frac{\sin \alpha}{a} = \displaystyle \frac{\sin \beta}{b}

\displaystyle \frac{\sin 30^\circ}{5} = \displaystyle \frac{\sin 60^\circ}{b}

\displaystyle \frac{1/2}{5} = \displaystyle \frac{\sqrt{3}/2}{b}

b = 5\sqrt{3}

green lineIn the next few posts of this series, I’ll consider the other SSA cases — including the case where two solutions are possible.