OA132
Trigonometry

Introduction

In the lectures and modules that you have seen so far, you will have seen a wide variety of x-y graphs. However, this is not the only way we can describe the distribution of data points, the relationship between straight lines, etc. We also need to examine the angle between lines and planes (e.g., in examining the dip and strike of geologic strata), we need to look at waves (most obviously in the oceans, but also related to seismic activity) and to look at spherical geometry (the Earth is not flat!). This approach requires us to understand something about the mathematics of angles, and is the subject of this module.

The Basics

Consider the right angled triangle below.

In this diagram the various lines are defined as: h = hypotenuse, a = adjacent and o = opposite. The relationship between the length of these lines and the angle X, is given by:

opposite/adjacent = o/a = tangent of x = tan x

adjacent/hypotenuse = a/h = cosine of x = cos x

opposite/hypotenuse = o/h = sine of x = sin x

You need to have a calculator that lists all of these functions and you need to know how to operate your calculator to obtain these functions and their inverse, e.g.,

sin 68^o = 0.92718 : sin^-1 0.92718 = 68^o

cos 12^o = 0.97815 : cos^-1 0.97815 = 12^o

tan 32^o = 0.62487 : tan^-1 0.62487 = 32^o

Check that you can do this on your calculator and that you get the same answers.

Most of you should be familiar with the fact that a circle is comprised of 360^o. However, just as we can measure distance in metres, yards, chains, etc., we can also measure parts of a circle in radians. A complete circle is comprised of 2p radians, hence one radian = 57.3^o.

On many calculators it is possible to choose whether to calculate trigonometric functions in either degrees or radians. You need to check whether you can do this on your calculator. If not, then you need to remember how to convert between radians and degrees, so that you can make the appropriate adjustment when it comes to making certain calculations.

Dips, Strikes and Bed Thickness

You are carrying out a mapping exercise and encounter the rocks below.

The rocks are dipping at angle of 39^o and they are exposed over a distance of 0.8 km (perpendicular to the strike). What is the true thickness of this unit?

The problem can be represented as a right-angled triangle in which the length of the hypotenuse is equivalent to the length of the exposure and the labelled angle is equivalent to the dip of the beds.

The length of the side of the triange opposite to the angle is the true thickness of the bed and is given by:

thickness = 0.8 x sin 39^o = 0.8 x 0.629 = 0.503 km = 503 m

If you are taking a geologic map into the field you may wish to relate dips on a map cross section to dips on the ground. Many cross sections on geologic maps have vertical exaggerations. In the example below this exaggeration is x10.

If the dip of the contact between the Triassic and Lias beds is 22^o on the cross section, what will it be on the ground? You might guess 2.2^o, but the relationship between angle and sine (a measure of the vertical scale) is not linear, as shown below.

The sine of 22^o is 0.3746. Divide this by 10 to account for the vertical exaggeration (i.e., 0.03746) and take the inverse sine (sin^-1) to get the true dip of 2.15^o.

More Advanced Rules

Of course, not all triangles contain a right angle. So, we need a set of rules to cover their trigonometry. Consider the triangle below.

There are three rules that define the relationship between the length of the triangle sides and the angles.

Sine rule:

a/(sinA) = b/(sinB) = c/(sinC)

Cosine rule:

a² = b² + c² - (2bc. cosA)

b² = a² + c² - (2ac. cos B)

c² = a² + b² - (2ab. cos C)

180^o rule:

A + B + C = 180^o

The combination of these rules allows us to calculate all 3 angles and 3 side lengths, providing we know 3 pieces of information. However, there is a small complication that must be borne in mind. The diagram below shows that between 0-180^o the same value of the sin can give two different values for the angle.

For this reason, it is a good idea to avoid the sine rule unless there are other means of defining the limits of a particular angle.

In this example, calculate the lengths of sides a and c and the size of angle B (note diagram is not to scale).

First use the 180^o rule.

25.6 + 106.3 + B = 180 : hence, B = 180 - 25.6 - 106.3 = 48.1^o

We can use the sine rule here as we know the angles. So,

a/sin106.3 = 1.65/sin48.1 : hence, a/0.9598 = 1.65/0.7443 : a = 1.65 x 0.9598/0.7443 = 2.13 km

c/sin 25.6 = 1.65/sin48.1 : hence c/0.4321 = 1.65/0.7443 : c = 1.65 x 0.4321/0.7443 = 0.96 km

In geology we are often confronted with a 2-dimensional representation of a 3-dimensional problem. For example, a flat cliff exposure only gives 2-dimensions.

As long as we know the strike of the beds (e.g., from an inland exposure of the same beds). We can calculate the true dip.

The green translucent plane represents the bedding plane. We view the cliff face from the bottom right, so that we see the plane apparently dipping with an angle q' (18^o in this case), defined by the triangle o-a-c.

If we look down on the cliff exposure from above we could define a right angled triangle o-b-c, where the angle a is the angle between the cliff face and the true strike (33^o in our example) of the bedding plane.

Finally, we can define a third right-angled triangle o-a-b, where the angle q is the true dip.

As we have measured we know,

tan q' = oa/oc = (oa/ob).(ob/oc) : (bottom right triangle)

as, ob/oc = cos a : (top triangle)

then, tan q' = (oa/ob).cos a

as, oa/ob = tan q : (bottom left triangle)

then, tan q' = tan q.cos a

which we can rearrange to:

tan q = tan q'/cos a : hence q = tan^-1(tan q'/cos a)

In our case q = tan^-1(tan18^o/cos33^o) = tan^-1(0.325/0.839) = tan^-1(0.387) = 21.2^o = true dip.

Waves

Trigonometric functions are also required in studying waves. First, a few basic definitions.

L (upper case lambda) is the wavelength

H is the total wave height and A = H/2

T is the time for one wavelength to pass a fixed point

h (lower case eta) is the wave height above or below the average height at a specified time and place

C (stands for celerity) is the wave speed

k = wave number

w (lower case omega) = frequency

L = CT

k = 2p/L

w = 2p/T

The wave height as a function of time and distance along the x-axis h(x,t), i.e., along the direction the wave is travelling, is given by:

h(x,t) = A cos(kx - wt)

where the angle is measured in radians, rather than degrees.

In the oceans, the relationship between wind speed and wave height and speed are dependent on a variety of factors, including the fetch (i.e., the distance the waves are able to travel before they meet land), the depth (i.e., whether the waves interact with the bottom) and the stability of the wind speed. However, in the open oceans wave speed (C) is generally ~1/3 of the wind speed (v), i.e.,

C = v/3

Similar empirical relationships have been developed between wave height (H) and wind speed, and wave length (L) and wave speed:

C = 1.25 L^0.5

H = 0.0071 v^2.5

Hence, we can apply these equations to a particular wind speed and calculate the various parameters they contain.

For example, consider a wind speed (v) of 10 ms^-1 (force 5, ideal for sailing).

C = v/3 = 3.33 ms^-1 : L = (C/1.25)² = 7.11 m : T = L/C = 7.11/3.33 = 2.133 s : H = 0.0071 v^2.5 = 2.25 m : A = H/2 = 1.12 m

We can now use a spreadsheet to calculate h as a function of x and t. First set up the spreadsheet to calculate the constants k and w.

There are programs that allow you to show h on a 3-dimensional diagram with axes of h, time (t) and distance (x), but we can only display h versus one of these variables at a time. So, if we need to fix either x or t. Fixing x = 0 allows us to examine how the wave height varies with time at one fixed point.

We can now start building the equation for h into the spreadsheet.

Now use Chart Wizard to plot h(x,t) = A cos(kx - wt) against t. (The range you use for t and the increments may require a bit of trial and error to ensure that you get a plot that adequately displays the function.)

Unfortunately, Excel does not allow you to a meaningful trendline through these points. If I had chosen more closely spaced values of t I could have used the moving average. Nevertheless, the wave like format of the plot should be readily apparent to you.

Spherical Trigonometry

In the examples we have considered so far, we have been able to ignore the fact that we live on a spherical planet and not a flat Earth. However, we have to deal with spherical geometry when we are dealing with large scale problems, such as the movement of tectonic plates.

The relative motion of two spreading plates on the surface of the Earth is similar to the opening of a hinge (see below).

In this case the axis of the hinge is the pole of rotation (P = green circle). Clearly, the further we are away from the axis of the hinge the faster the relative motion of the two sides of the hinge. So, at point x' the two sides of the hinge are moving apart more slowly than they are at point x''. However, the angular velocity is constant all the way along the centre of the opening hinge (the line from P to x'').

If you are having problems understanding this, think of an opening door. If takes you 2 seconds to open a door so that it is pointing at 90^o to the wall, then the door handle will have travelled further in the same time (i.e., faster) than a point on the door closer to the hinge. However, all points on the door will have opened by 90^o in the same time.

If we know the radius of the Earth, the angular velocity of two plates and the position of their pole of rotation we can calculate the relative plate motions at any point along their boundary. Consider the example below.

The green dot gives the position of the pole of rotation between the convergent African and South American plates (62.5^oN, 39.4^oW).

The red dot gives the position of spot on the mid-Atlantic ridge that forms the boundary between the two plates (7.1^oS, 15.3^oW).

The angular velocity between plates is 3.2 x 10^-7 deg yr^-1.

The various parameters needed to calculate the relative plate velocities and direction of movement at the red dot, together with the symbols are given in the table below. (The conventions are that ^oN and ^oE are positive and ^oS and ^oW are negative).

a = cos^-1[sin(lx).sin(lp) + cos(lx).cos(lp).cos(fp - fx)]

= cos^-1[sin(-7.1).sin(62.5) + cos(-7.1).cos(62.5).cos(-39.4 - -15.3)]

= cos^-1[(-0.1236 x 0.8870) + (0.9923 x 0.4617 x 0.9128)]

= cos^-1[(-0.1096) + (0.4182)] = 72.03^o

C = sin^-1[(cos(lp).sin(fp - fx)/sin(a)]

= sin^-1[(cos(62.5).sin(-39.4 - -15.3)/sin(72.03)]

= sin^-1[(0.4617 x -0.4083)/0.9512]

= sin^-1[-0.1885/0.9512] = -11.43^o

v = w x R x sin a

= 5.59 x 10^-9 x 6371 x sin(72.03) km yr^-1

= 3.39 x 10^-5 km yr^-1 = 3.39 cm yr^-1

b = 90 + C = 90 + -11.43 = 78.57^o

Hence, the plates are spreading apart at a rate of 3.39 cm yr^-1, and the spreading is slightly anti-clockwise from an East-West direction.

Reflection and Refraction

We all well aware of the phenomenom of reflection. However, an examination of the geometry and trigonometry of reflection is a useful introduction to the more complex process of refraction.

We can consider light hitting a mirror as a series of moving wave fronts, like waves washing up on to a beach. The wave front moves from A-B to A'-B', where it hits the reflecting surface. The waves hit the surface at angle of q₁, the incident angle. This angle is measured as the deviation from a line perpendicular to the surface of the mirror.

The leading edge of the wave hits the mirror and is reflected along the path A'-S. By the time it reaches the point S, the outer edge of the wave has struck the mirror at point R. The angle of reflection if given by q'₁.

If we consider the two triangles A'SR and A'B'R seperately, we can examine the relationship between the angles of incidence and reflection.

These triangles are termed "similar". They are not identical as they are mirror images of one another. We know that the angles A'-S-R and A'-B'-R are both right angles. Equally, we know that the length of side A'-S is equal to B'R and the side A'-R is common to both triangles. Hence, it follows from the basic definitions at the start of the module that angle S-R-A' (q'₁) is equal to angle R-A'-B' (q₁). In other words, the angle of incidence is equal to the angle of reflection. This seemingly obvious observation is termed "Huyghen's principle".

Let's now consider the more complex phenomenom of refraction. Refraction occurs when a wave passes from a medium of one density into another. This causes a change in the velocity of the wave and a deviation in its path. Common examples of refraction occur when the wave is light (e.g., when we look at an object below water it appears closer than is truly the case), a seismic wave (which is bent by passing between rocks of different density), or a water wave (which is bent towards the beach as it passes into shallow water). The basic principles are the same in all cases and are illustrated below.

In this case the velocity of medium 2 (V₂) is greater than that of medium 1 (V₁) so the wave is deflected to a shallower angle. The angle of refraction is given by q₂. We can again consider two triangles.

We can express the length of side A-S as equal to the sine of angle q'₁ times the length of side A'-R, i.e., A'S = A'R sin (q'₁) = A'R sin (q₁) (remember q'₁ = q₁). Similarly, from the bottom triangle we obtain, A'T = A'R sin (q₂). As A'R is common to both triangles, we can write:

A'S/ sin (q₁) = A'T/ sin (q₂)

and as A'S is proportional to the velocity of medium 1 (V₁) and A'T is proportional to V₂, it therefore follows that:

V₁/ sin (q₁) = V₂/ sin (q₂) or,

V₁/ V₂ = sin (q₁)/ sin (q₂)

This relationship is known as Snell's law.

A common application of the laws of refraction is in carrying out seismic surveys. At its simplest, this involves hitting the ground with a sledgehammer and deploying a series of recording devices at varying distances from the point where you hit the ground.

If we hit the ground at point S, one wave (the direct wave) will travel directly along the surface of the ground. The speed of this wave will depend on the velocity of the medium (i.e., V₁).

Another wave (the head wave) will through the surface medium (which has a thickness h). The wave will radiate out in all directions. The portion of the wave that hits the second medium at the critical angle (i_c) will be refracted so that it travels horizontally with a velocity V₂.

As the angle of refraction = 90^o, then sin (q₂) = 1 and from Snell's law V₁/ V₂ = sin (q₁). As this wave travels along the path A-B parts of the wave are refracted back up through the top medium at the same critical angle, i_c.

If we consider just one of those recorders (at point G), then the time taken for the direct wave to travel from S-G is given by:

t_d = SG/V₁

and the time taken by the refracted wave to travel from A-G is given by:

t_r = SA/V₁ + AB/V₂ + BG/V₁

We can measure t_d, t_r, x and V₁. We want to determine V₂ (as this gives us an idea of the composition of medium 2) and h (as this tells us how far medium 2 is below the surface).

h = SA cos (i_c), so SA = h/cos (i_c)

and as SA = BG then;

t_r = 2h/V₁cos (i_c) + AB/V₂

Now we can look at the triangle at the left end of the path SABG.

Opposite/adjacent = tan, so ab/h = tan (i_c). Hence, ab = h tan (i_c). The same applies to the triangle at the other end of SABG. Hence, the distance A-B is given by:

A-B = x - 2h tan (i_c), and

t_r = 2h/V₁cos (i_c) + (x - 2h tan (i_c))/V₂

we can now set about solving this equation in terms of the parameters we can measure.

we know that tan (i_c) = sin (i_c)/cos (i_c), so

now group the 2h/cos (i_c) terms to obtain,

now multiply the term inside the bracket by V₁ and divide the term outside of the bracket by V₁ to obtain,

remember from Snell's law (see above) that V₁/ V₂ = sin (q₁), so we can substitute sin (i_c) for V₁/ V₂ to obtain,

from Pythagorus's theorum we know that the sum of the square of the opposite and adjacent sides of a right angled triangle are equal to the square on the hypoteneuse.

So for, 1² = cos² + sin². Hence 1 - {sin (i_c)}² = {cos (i_c)}², and,

the cos (i_c) terms then cancel out to give,

if we return to Pythagorus's theorum we know that 1 - {sin (i_c)}² = {cos (i_c)}², and that V₁/ V₂ = sin (q₁), so

1 - ( V₁/ V₂)² = {cos (i_c)}² and cos (i_c) = [1 - ( V₁/ V₂)²]^0.5 Hence,

This equation has the form of y = mx + c; i.e., a straight line with an intercept. In this case a plot of t_r versus x will yield a straightline with a slope of 1/V₂ and a intercept equal to;

intercept =

Once you have solved the slope equation for V₂ you can solve the intercept equation for h.

Test

1. A church tower lies on a bearing of 63^o from where you are standing. At a bearing of 206^o there is a pub. You know it is 1.93 km from the pub to the church and 1.62 km from you to the church. How far are you from the pub?

Not sure to do?

Hint 1

Still having problems?

Hint 2

You should end up with a plot that looks something like this.

Answer 1

2. Under the waves section of this module I showed the case of plotting wave height (h) versus time. Using the same information, construct a spreadsheet to calculate h versus distance (x) and constant time (t = 0).

Hint 3

You should end up with a plot that looks something like this.

Answer 2