To the Stars

Maths can be a dry subject to learn when you do not know of its many applications.This post will be about one of the many applications of trigonometry. Trigonometry is used extensively in the GPS system to determine your position. But I’d like to go a bit farther out in space and show how distances to stars are determined.

Let me first set up the example and then define some new measurement units before I actually develop the solution. This is because of the large distances and small angles used in astronomy.

A common method to calculate the distance of nearby stars is to use parallax. You can see parallax in action by stretching out your hand and extend your index finger, then alternately close one eye and see how your finger changes position compared to objects further away. This effect is rather pronounced with your finger because the distance between your eyes is not much smaller than the distance to your finger.

What astronomers do is to see the relative position of of a nearby star at different places compared to farther stars that do not change their position much over time. But any distance on earth is very small compared to the distance to a star so the effect is too small to measure. What to do?

Well as we orbit our star (the sun), we could make a measurement on one side of the orbit, say in June, then on the other side in December. This would be a much larger distance than anything we could do on earth and the resulting parallax effect would be measurable. So the set up of the problem looks like this:

So the two readings are needed to determine the angle 𝜃. Once 𝜃 s known, we can use it to calculate d. From the previous post on trig functions, you can see that for this problem:

\[\tan\hspace{0.33em}\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{r}{d}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{d}\hspace{0.33em}{=}\hspace{0.33em}\frac{r}{\tan\hspace{0.33em}\mathit{\theta}}\]

But there are some practical problems here. Even though we are using the largest possible distance while restricted to the earth’s surface to measure the parallax angle, this angle is still very very small. So astronomers use arc seconds (arcsec). This is 1/3600 of a degree. Also, as distances to stars are very large, the unit of distance between the earth and the sun used is 1 AU (astronomical unit). 1 AU equals 149.6 million kilometers. Using these units results in the distance expressed in parsecs (yes, parsecs is a distance measurement, not a time measurement as used in the Star Wars movies). However, we will just use AU’s for distance and degrees for the angle in the example.

So how far is a star with a measured parallax angle of 0.0002 degrees?

\[{d}\hspace{0.33em}{=}\hspace{0.33em}\frac{{1}\hspace{0.33em}{AU}}{\tan\hspace{0.33em}{(}{0}{.}{0002}{)}}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{0.00000349065}\hspace{0.33em}{=}\hspace{0.33em}{286479}{.}{6}\hspace{0.33em}{\mathrm{AU}}\]

You can see why astronomers use arcsec. Using the all knowing Google, converting 286479.6 AU gives 42,900,000,000,000 km or 4.53 light years or 1.39 parsecs.

What’s Your Sine?

Continuing with the trigonometry theme, there are other relationships between the angles and the length of the sides of a right triangle. You may have a calculator with buttons labeled as “sin” or “sin x” or “sin θ”. There would be similar buttons using the prefix “cos” and “tan”. These are the basic trig (trigonometric, hence the abbreviation) functions. What do these functions do?

As with the Pythagorus theorem covered in my last post, there are other relationships that apply to right triangles regardless of their size. But unlike the Pythagorus theorem which relates the lengths of just the sides, the trig functions relate the side lengths with the internal angles.

But before I show these, remember that there are several ways to measure angles, degrees and radians being the most common. When using the trig functions, your calculator needs to know what measurement you are using. As we have been and will continue to use degrees, you need to make sure that in your calculator settings, you have set the degrees mode. In most calculators, this will show up as an abbreviation “deg”. Radians will display as “rad”. As a full circle angle is 360° and approximately 6.28 radians, there is quite a bit of difference between the two types of measurements. So let’s begin.

So here is your basic, everyday right triangle with the angle of interest, 𝜃, and the sides labelled with respect to that angle. The side could be adjacent to it or opposite it. The hypotenuse, as seen before, is the side opposite the right angle. For any size right triangle, the basic trig functions are defined as follows:

\[
\sin\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{\mathrm{hyp}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\cos\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{adj}}{\mathrm{hyp}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\tan\mathit{\theta}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{\mathrm{adj}}\hspace{0.33em}{=}\hspace{0.33em}\frac{\sin\mathit{\theta}}{\cos\mathit{\theta}}
\]

So the trig functions show the ratios of the various sides. In the above formulas, the following abbreviations are used: sin = sine, cos = cosine, tan = tangent, opp = opposite, adj = adjacent, hyp = hypotenuse. If you type 45 into your calculator and hit the “sin” key, you should see a decimal number 0.7071… . This means that for a right triangle with a 45° angle, the length of the opposite side divided by the length of the hypotenuse is always 0.7071…

The power of these functions is that you only need the length of one side and an angle (other than the right angle) of a right triangle to determine the lengths of the other two sides.

Example:

What are the lengths of the other two sides of the above triangle?

\[
\begin{array}{l}
{\sin{30}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{5}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{5}}\\
{{\mathrm{opp}}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{5}\hspace{0.33em}\times\hspace{0.33em}{5}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{5}}\\
{\cos{30}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{adj}}{5}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{866}}\\
{{\mathrm{adj}}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{866}\hspace{0.33em}\times\hspace{0.33em}{5}\hspace{0.33em}{=}\hspace{0.33em}{4}{.}{33}}
\end{array}
\]

So the lengths are 2.5 and 4.33. The units (centimeters, inches, etc) are whatever the units of the given hypotenuse is.

Let’s try this on a practical problem:

You are 500 meters from a tall tower. Using a sextant (a device to measure angles), you measure the angle between the ground and the line between your feet and the top of the tower to be 15°. How high is the tower?

The side adjacent to the angle is 500 meters. We need to find the side opposite (the tower). Looking at the trig functions, the tangent looks like it is the best one to use since it is the ratio of the side opposite (unknown) and the side adjacent (known). Finding the tangent of 15° on my calculator gives 0.2679. So,

\[
\begin{array}{l}
{\tan\hspace{0.33em}{15}\hspace{0.33em}{=}\hspace{0.33em}\frac{\mathrm{opp}}{\mathrm{adj}}\hspace{0.33em}{=}\hspace{0.33em}\frac{{\mathrm{tower}}\hspace{0.33em}{\mathrm{height}}}{{\mathrm{distance}}\hspace{0.33em}{\mathrm{from}}\hspace{0.33em}{\mathrm{tower}}}\hspace{0.33em}}\\
{{=}\frac{{\mathrm{tower}}\hspace{0.33em}{\mathrm{height}}}{500}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{2679}}\\
{{\mathrm{tower}}\hspace{0.33em}{\mathrm{height}}\hspace{0.33em}{=}\hspace{0.33em}{0}{.}{2679}\hspace{0.33em}\times\hspace{0.33em}{500}\hspace{0.33em}{=}\hspace{0.33em}{133}{.}{97}\hspace{0.33em}{\mathrm{meters}}}
\end{array}
\]

Now that saves a lot of effort to physically measure the height, doesn’t it?

The Correct Angle is the Right One

In my last post, you saw that a 90° angle is called a right angle. This is the angle made by the two lines at the corner of a square. Now a triangle is a shape that has three angles inside. A basic property of any triangle is that all the internal angles add up to 180°:

But this post is about triangles where one of its angles is 90°, that is a right angle. Such triangles are called right triangles.

Below is a right triangle where I have labelled the sides as a, b, and c. Side c is the side opposite the right angle. This side is called the hypotenuse. The hypotenuse is always the longest side of any right triangle.

Right triangles have another famous property that relates the lengths of the three sides. This property is called the Pythagorus Theorem. This is named after the Greek mathematician Pythagorus who lived 570 to 495 BCE. This theorem was used before his time but he is credited with providing the first proof. Given the sides as labelled above, the following is true for any size right triangle:

c² = a² + b²

This means that if you know any two sides of a right triangle, you can calculate the third side using this equation. Let’s do an example:

So we now know that c² = 4² + 3² = 16 + 9 = 25. You can now find c by taking the square root of both side of the equation. Square roots have been covered in previous posts:

\[{c}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}\sqrt{{c}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{25}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{c}\hspace{0.33em}{=}\hspace{0.33em}{5}\]

In my posts on square roots, I did say that taking a square root results in two solutions, one positive and one negative. But since we are solving for a physical length, we can ignore the negative solution. So the hypotenuse for this triangle is 5.

Not all right triangle problems work out so well. Most square roots are decimal numbers and you have to either round the number or leave the answer as a square root.

The Pythagorus Theorem can also be used to find one of the non-hypotenuse sides as well:

You can rearrange the theorem’s equation to solve for the unknown side:

\[\begin{array}{l}{{c}^{2}\hspace{0.33em}{=}\hspace{0.33em}{a}^{2}\hspace{0.33em}{+}\hspace{0.33em}{b}^{2}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{b}^{2}\hspace{0.33em}{=}\hspace{0.33em}{c}^{2}\hspace{0.33em}{-}\hspace{0.33em}{a}^{2}}\\{{b}^{2}\hspace{0.33em}{=}\hspace{0.33em}{9}^{2}\hspace{0.33em}{-}\hspace{0.33em}{5}^{2}\hspace{0.33em}{=}\hspace{0.33em}{81}\hspace{0.33em}{-}\hspace{0.33em}{25}\hspace{0.33em}{=}\hspace{0.33em}{56}}\\{\sqrt{{b}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{56}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{b}\hspace{0.33em}\approx\hspace{0.33em}{7}{.}{48}}\end{array}\]

So b approximately equals 7.48. That is what the symbol “≈” means. The exact answer cannot be written as a decimal number as the decimal part goes on forever.