August 21, 2019, 11:32:04 pm

### AuthorTopic: The sine rule's mysterious *Ambiguous* case  (Read 84 times)

0 Members and 1 Guest are viewing this topic.

#### RuiAce

• ATAR Notes Lecturer
• Moderator
• Great Wonder of ATAR Notes
• Posts: 8448
• "All models are wrong, but some are useful."
• Respect: +2312
##### The sine rule's mysterious *Ambiguous* case
« on: July 19, 2019, 07:20:45 pm »
+5

Using the sine rule generally requires us being cautious when it comes down to finding an angle, instead of a side.

Background
Somewhat surprisingly, using the sine rule for the sake of finding an angle is one of the most dangerous things a mathematician/physicist/engineer/... could ever have to do. The sine rule doesn't always behave well, because of how the trigonometric function $\sin$ itself works.

To motivate this, let's consider the following (arguably innocent) equation. We'll do it in degrees.
$\text{Solve }\sin \theta = \frac{1}{\sqrt2}\text{, where }0^\circ< \theta <180^\circ.$
This is telling us to look for solutions in the first and second quadrant, both of which we know $\sin$ is positive in. So we'd be inclined to say $\theta = 45^\circ$, or $\theta =1 80^\circ - 45^\circ =135^\circ$.

The problem: We have two distinct solutions. How do we know which is the correct one?

In general, the sine rule (in finding an angle) will always boil down to the issue of solving $\sin \theta = \text{'something'}$. And because of how the $\sin$ function behaves, we always get two angles. The whole point of stating $0^\circ < \theta < 180^\circ$ was to iterate the fact that angles in a triangle are always between zero and 180 degrees. (Of course, assuming that it's not degenerate, but we won't talk about that.)

So do we ever know which one will be the correct one?
Let's consider the scenario. The objective is to find $\theta$. We're given that $AB=5$, $AC=8$ and $\angle B = 70^\circ$.
(Note: The sides weren't labelled here.)
We certainly have enough information to apply the sine rule, so let us do it.
\begin{align*}
\frac{8}{\sin 70^\circ} &= \frac{5}{\sin \theta}\\
\sin\theta &= \frac{5\sin 70^\circ}{8}\\
\theta &\approx 35^\circ 58', \, 144^\circ 2'
\end{align*}
...oh right, because $\theta$ could be obtuse, and therefore it could be a second quadrant angle. So we actually needed to do something along the lines of ASTC thinking here.

So which one is the correct one?

Yeah look, if you assume that my diagram is correct, then you might say "well duh $\theta$ is acute in that picture, so it will be (approx) $35^\circ 58'$." But what if my diagram was all a lie and in fact $144^\circ 2'$ was the correct value and we didn't realise that?

Here's a better example, taken from the HSC Cambridge textbook.

As you can see, there may as well be another triangle that we've completely ignored in the whole process. The painful truth is that we can't always work around this issue! So here's what will happen.
____________________________________________________
1. QCAA may tell you in words whether or not the angle is obtuse or acute. That basically says "they've made the decision for you, now you obediently listen to them."

2. The angle sum of a triangle being $180^\circ$ enforces a cap. In the example I produced, suppose that the angle we required was $144^\circ 2'$. But we already know one of the angles is $70^\circ$, so together these would give us an angle sum of $214^\circ2'$, which is definitely more than what's allowed for a triangle! So in that example, we can justifiably eliminate the larger angle and leave us with only $35^\circ 58'$.

3. This method is mostly inspired by Eddie Woo in this specific video. You have to watch pretty far into it to see him explain it. But the intuition is the following.

In general, the larger the angle, the longer the opposite side of the triangle ends up being. This essentially occurs because making one particular angle larger in a triangle, whilst holding its two neighbouring sides at constant size, requires us to stretch out the opposite side at the same time.

This gives a useful consequence. If we already know one angle-and-opposite-side pair, if that side is longer than the side corresponding to our $\theta$, that angle must also be larger than our $\theta$ as well. And also vice versa. This sometimes allows us to eliminate one of the angles.
____________________________________________________

Of course, points 2 and 3 are not fool-proof - they're just viable workarounds. Point 1 is guaranteed to work if QCAA does specify it, but at this stage who knows if they will. So it's definitely worth keeping those thoughts at the back of your head to help resolve this issue!

#### laura_

• MOTM: JUNE 19