Category: Notes

 

Using the Law of Sines

In any triangle $\mathrm{\Delta }ABC$, it will be true that $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

To use it, we must already know three out of four of the pieces of information. So we need to have one of the following situations:

• We know one angle and the length of its opposite side, and we know the angle opposite the side we want to find. (This is AAS from the congruence theorems of geometry. In case we have ASA, we can find the third angle and then use the Law of Sines.)
• We know one angle and the length of its opposite side, and we know the length of the side opposite the angle we want to find. (This is an ambiguous case and care must be taken in using the Law of Sines in this case: see at the end of the Law of Cosines example below.)

Notice that in both cases, we need to know one angle and the length of that angle’s opposite side. If we don’t have that, the Law of Sines cannot be used.

Example:

In $\mathrm{\Delta }ABC$, $\mathrm{\angle }A={30}^{\circ }$, $\mathrm{\angle }B={50}^{\circ }$, and $a=5$. Find the length of side $b$.

For this problem we will use the first two ratios in the Law of Sines:

$\frac{\sin A}{a}=\frac{\sin B}{b}$

Substituting in the given information:

$\frac{\sin {30}^{\circ }}{5}=\frac{\sin {50}^{\circ }}{b}$

Now solve for $b$ (do it algebraically before you compute any numbers!)

$b\sin {30}^{\circ }=5\sin {50}^{\circ }$

$b=\frac{5\sin {50}^{\circ }}{\sin {30}^{\circ }}$

Now enter that last expression into your calculator: first check that you are in degree mode!

$b\approx 7.66$

Continuing the Example:

We can now go on to solve this triangle: the remaining unknowns are $c$ and $\mathrm{\angle }C$. Remember, this is not a right triangle, so we cannot use the Pythagorean Theorem. But we can use the fact that the sum of all the angles in the triangle must be ${180}^{\circ }$ to find $\mathrm{\angle }C=\text{1}{80}^{\circ }-({30}^{\circ }+{50}^{\circ }={100}^{\circ }$, and then use the Law of Sines again to find the length $c$. i There are two ways to do this, but since we had to round $b$, it is better to use $a$ and $\mathrm{\angle }A$ in our proportion:

$\frac{\sin A}{a}=\frac{\sin C}{c}$

$\frac{\sin {30}^{\circ }}{5}=\frac{\sin {100}^{\circ }}{c}$

Solving this proportion as before, we get

$c=\frac{5\sin {100}^{\circ }}{\sin {30}^{\circ }}$

Enter this in the calculator to compute:

$c\approx 9.85$

—————————————————-

We can also use the Law os Sines to find an angle. That can be problematic, though. We will do that next time if time permits.

—————————————————-

Using the Law of Cosines:

For any triangle $\mathrm{\Delta }ABC$, the following are true:

$a^2=b^2+c^2–2bc\cos A$

$b^2=a^2+c^2–2ac\cos B$

$c^2=a^2+b^2–2ab\cos C$

[Notice that if $C$ is a right angle, so $\cos (C)=0$, the last identity is just the Pythagorean Theorem.]

Using them:

The Law of Cosines is especially useful when we know the length of two sides and their included angle. (This is SAS from the congruence theorems of geometry.) It is also our only choice to use for finding an angle in the SSS case. However, care must be taken in that case!

Example:

If $a=13$, $b=12$ and $\text{Extra close brace or missing open brace}$, find the length of side $c$.

We use the third formula above:

$c^2=a^2+b^2–2bc\cos C$

$c^2={13}^2+{12}^2–2(13)(12)\cos {78}^{\circ }$

Don’t compute anything yet, just solve for $c$:

$c=\sqrt{{13}^2+{12}^2–2(13)(12)\cos {78}^{\circ }}$

Only the positive square root is needed here because $c$ is a length.

Now enter that into your calculator to find $c$:

$c^2={13}^2+{12}^2–2(13)(12)\cos {78}^{\circ }$75

$c\approx 15.75$

Continuing this Example:

We can now go on to solve this triangle. Remember that it is always best to avoid using rounded numbers in our computations, and when forced to do so we should include some extra digits in the decimal part.

We now know all three sides of the triangle and the measure of $\mathrm{\angle }C$. We need to find the measures of the other two angles. Notice that once we know one of them, the other will be easy to find.

Here is one thing we could do:

Using sides $a$ and $c$ and angle $C$, we could use the Law of Sines to find $\mathrm{\angle }A$. Since we only have $c$ as a rounded number, we will do one of two things (depending on what device/program we are using to calculate):

• Compute $c$ to a few more places, let’s say to 6 places at least. (The more the better.)

• Don’t compute $c$ at all, but enter the expression using the square root into our formula. This is the best way, but if doing it on a hand-held calculator may be hard to enter.

I will show both ways.

We use this version of the Law of Sines:

$\frac{\sin A}{a}=\frac{\sin C}{c}$

$\frac{\sin A}{13}=\frac{\sin {78}^{\circ }}{c}$

Before we substitute for $c$ in that formula, we need a few more decimal digits, so go back and compute:

$c\approx 15.752192$ to six decimal places

Substitute this in and solve for $\sin A$:

$\frac{\sin A}{13}\approx \frac{\sin {78}^{\circ }}{15.752192}$

$\sin A\approx \frac{13\sin {78}^{\circ }}{15.752192}$

Don’t compute anything else yet! We want $\mathrm{\angle }A$, not $\sin A$. So let’s compute

${\sin }^{-1}\left(\frac{13\sin {78}^{\circ }}{15.752192}\right)\approx {52.83}^{\circ }$

This may or may not give us $\mathrm{\angle }A$. In this case, since the angle is positive, we can be assured that we did in fact get the correct angle. The problem arises when ${\sin }^{-1}$ gives a negative angle: then we will have to correct it to the actual angle in the triangle.

So $\mathrm{\angle }A\approx {52.83}^{\circ }$.

We can now use the fact that all the angles in the triangle add up to ${180}^{\circ }$ to find an approximation for angle $B$: 

$\mathrm{\angle }B\approx {180}^{\circ }–({78}^{\circ }+{52.83}^{\circ })$

$\mathrm{\angle }B\approx {49.17}^{\circ }$

We have now solved the triangle.

——

Finding angle $B$ without approximating $c$:

We would do the same as above, but instead of substituting in the approximation $c\approx 15.752192$, we substitute in the formula $c=\sqrt{{13}^2+{12}^2–2(13)(12)\cos {78}^{\circ }}$:

$\sin A\approx \frac{13\sin {78}^{\circ }}{\sqrt{{13}^2+{12}^2–2(13)(12)\cos {78}^{\circ }}}$}$

$\text{Missing or unrecognized delimiter for left}$}$

Enter this in your calculator – easier to do if you are using Desmos scientific calculator or something like that! – and find that $\mathrm{\angle }A\approx {52.83}^{\circ }$ to two decimal places, just as in the other way.

As long as you include enough decimal places in the numbers you enter into your calculations, the final result will be the same after rounding. But it’s always better to avoid using rounded numbers in calculations if you can.

This example is intended to show you what not to do when solving rational equations. Not because it is wrong – everything done here is correct – but because it makes the problem much harder than it needs to be.

Your work will always be easier if you clear the denominators in the equation by multiplying by the LCM of all the denominators. So practice finding the LCM if necessary. Here is a Khan Academy video example.

 

This is one of the versions of problem 4 on Test 3. To see a solution which avoids the complexities of this, look at the Test 3 solutions

Solve $3+\frac{9}{x-3}=\frac{27}{x^2–3x}$

Suppose my first thought is to combine the two terms on the left-hand side. Nothing wrong with that, provided I change to a common denominator first.

$3+\frac{9}{x-3}=\frac{27}{x^2–3x}$

$\frac{3(x-3)}{x-3}+\frac{9}{x-3}=\frac{27}{x^2–3x}$

$\frac{3x–9}{x-3}+\frac{9}{x-3}=\frac{27}{x^2–3x}$

$\frac{3x}{x-3}=\frac{27}{x^2–3x}$

And now, instead of clearing the denominators using the LCM, I will just cross-multiply. Nothing wrong with that either. Cross-multiplying is the same as clearing the denominators, except that instead of the LCM it (covertly) uses the product of the denominators.

$\text{Missing left or extra right}$

$3x^3–9x^2=27x–81$

$3x^3–9x^2–27x+81=0$

This is now a thrid-degree (cubic) equation! They can be very hard to solve in general. If I am lucky, this can be solved by factoring and using the Zero Product Property. Let’s see.

I notice that I can divide the whole equation by 3, and get

$x^3–3x^2–9x+27=0$

The only technique of factoring that we have learned in this course which applies to a polynomial with four terms is factoring by grouping. It does not always work, but let’s try it.

Group:

$\left(x^3–3x^2\right)–(9x–27)=0$

Factor out the common factor in each group:

$x^2\left(x–3\right)–9(x–3)=0$

And how do I know that my plan will work?* Because the factors in parentheses are the same! (If they were not the same, I could not factor this way.)

$\left(x^2–9\right)\left(x–3\right)=0$

factor some more:

$\left(x–3\right)(x+3)\left(x–3\right)=0$

Two of those factors are the same, so the Zero Product Property says that

either $x-3=0$ or $x+3=0$

$⟹x=3$ or $x=-3$.

But I have to check that these solve the original equation. $x=3$ gives a zero denominator,

so the only solution is $x=-3$

 

The complication could have been avoided by using the LCM of the denominators.

 

* At this point, I said, of course this reminds me of something, everything reminds me of something, and everything reminds me of my dog even though I don’t have a dog. Can you guess what I was reminded of? Hint: Hamilton, as usual.

First, for the myth debunking, a/k/a “When a student tells you one thing and your teacher tells you a different thing, which one is more likely to be right?’

We worked on this problem in class:
${\left(\frac{5a^4{z}^{-7}}{3a^{-6}z-4}\right)}^{-3}$
The student who was working it at the board started out by trying to simplify what is inside the parentheses first, which, I would like to point out, is absolutely correct: however, she made a mistake in that simplification. When I stopped her and said there was an error, several students told her that she was “supposed” to change the -3 power into the 3rd power of the reciprocal. (Not necessarily in those words.) I then countered by saying as insistently as I could that no, she did not have to do that first, she could simplify inside the parentheses first, I was only objecting that she had made an error in simplifying.

So what happened? The student listened to the students, who were WRONG in what they said.

It would have been totally correct to simplify inside the parentheses first and then deal with the -3 power. As I will do here:

${\left(\frac{5a^4{z}^{-7}}{3a^{-6}z-4}\right)}^{-3}={\left(\frac{5a^4{a}^6{z}^4}{3z7}\right)}^{-3}$
$={\left(\frac{5a^{10}{z}^4}{3z7}\right)}^{-3}$
$={\left(\frac{5a^{10}}{3z3}\right)}^{-3}$
$={\left(\frac{3z3}{5a^{10}}\right)}^3$
$=\frac{27z9}{125a^{30}}$

 

Morals of the story: there are at least two!

• When a student says one thing and the teacher says the opposite, maybe it’s better to listen to the teacher?

• There is not only one correct way to work these problems, and, indeed, that is true of a lot of what we do in this course. The important thing is to know the correct properties and definitions and how to use them correctly: they are your tools. And then, in working a problem, you look at what you have at each step and ask, what can I do with this? What tools can I use that may get me where I want to go? As long as you use mathematically correct “tools” and use them correctly and get to what the problem was asking for, your work is correct.

Thus it is your job to look to the examples to understand why we choose to do what we do at each step: what in the problem itself is suggesting to use this particular tool? Avoid at all costs memorizing “steps” whose purpose you do not understand. That is the opposite of mathematics!