Applications of trigonometry
Stand in Dublin and the Earth's spin carries you round at about 1,000 km/h; stand on the equator and it is about 1,670 km/h. Why the difference? The Earth is (almost) a sphere and its surface is described with two angles, latitude and longitude. We will use the cosine to work out how big a parallel of latitude really is, one of the horizontal lines on a map, and the distance between two places on the same one.
The radius of a parallel: why the cosine appears
A parallel is a circle made up of all the points on the Earth that are at the same latitude (the angle measured from the plane of the equator). The equator, at latitude $0^\circ$, has the same radius as the whole Earth; any other parallel is smaller, because it "narrows" as it gets closer to the poles.
Take a point P on the parallel and look at the vertical cross-section through P and the Earth's axis. The latitude $\alpha$ is the angle at the centre O of the Earth between the radius OP and the equatorial plane (the horizontal radius). Drop a perpendicular from P to the equatorial plane, meeting it at F. The right angle is at F. In triangle OFP the hypotenuse is $|OP| = R$, the angle at O is $\alpha$, and the side $|OF|$ adjacent to $\alpha$ equals $r$, the radius of the parallel (it is the same length as the distance from P to the axis).
For a school project, you need to calculate the length of the parallel of latitude $40^\circ$ N. Knowing that the Earth's radius is $R \approx 6\,371$ km, calculate: a)…
For a school project, you need to calculate the length of the parallel of latitude $40^\circ$ N. Knowing that the Earth's radius is $R \approx 6\,371$ km, calculate:
- a) The radius of the circle formed by this parallel (use the expression $r = R \times \cos(\text{latitud})$ and justify it from the right-angled triangle in the figure).
- b) The total length (circumference) of this parallel.
The Earth's equator has an approximate perimeter of $40\,030\text{ km}$ ($2\pi R$, with $R \approx 6371\text{ km}$). A cartography team needs to identify along which…
The Earth's equator has an approximate perimeter of $40\,030\text{ km}$ ($2\pi R$, with $R \approx 6371\text{ km}$). A cartography team needs to identify along which parallel in the Northern Hemisphere the radius of the parallel's circumference is exactly half of the Earth's radius ($r = \frac{R}{2}$).
- a) Set up a trigonometric equation relating the latitude $\alpha$ to this geometric condition using the formula $r = R\times\cos\alpha$, simplifying the radius $R$.
- b) Without using a calculator, deduce with reasons the exact latitude $\alpha$ at which this occurs, recalling the trigonometric values of special angles ($30^\circ$, $45^\circ$, $60^\circ$).
- c) What is the relationship between the total length of this parallel and the perimeter of the equator? Justify your answer without needing to calculate the exact kilometers.
The length of a parallel
Once the radius $r$ of a parallel is known, calculating its length (its perimeter) uses the same formula as always for a circle: $2\pi r$. The difference from the equator is that here the radius is no longer $R$, but the $r$ reduced by the cosine of the latitude.
Speed of rotation
The Earth turns once in $24$ hours, so a point on a parallel travels once round that parallel in $24$ h. Its speed is the length of the parallel divided by $24$.
The city of Dublin is located approximately at a latitude of $\alpha = 40^\circ\text{ N}$. Due to the Earth's rotation, anyone in Dublin describes a complete circle…
The city of Dublin is located approximately at a latitude of $\alpha = 40^\circ\text{ N}$. Due to the Earth's rotation, anyone in Dublin describes a complete circle corresponding to their parallel every $24\text{ hours}$. Consider the mean radius of the Earth to be $R = 6371\text{ km}$.
- a) Before using the calculator: knowing that at the equator (where the Earth's perimeter is around $40\,000\text{ km}$) the linear speed is about $1670\text{ km/h}$, and recalling that $\cos 45^\circ \approx 0.71$, qualitatively estimate whether the linear rotation speed in Dublin will be greater than or less than $1150\text{ km/h}$. Explain the reasoning behind your estimate.
- b) Accurately calculate the radius $r$ of the parallel of Dublin, the length of that parallel, and the actual linear speed in $\text{km/h}$ with which the city moves through space (use $\cos 40^\circ \approx 0.766$).
- c) Check whether the obtained value confirms your initial estimate and interpret what would happen to that linear speed if we moved toward the North Pole ($90^\circ\text{ N}$).
Distance between two points on the same parallel
If two cities are on the same parallel but at different longitudes (the east–west angle), the distance between them along the parallel is an arc of that smaller circle, not of the whole globe. You first need the radius of the parallel, and then the usual arc length formula with the angle that separates them.
For a school project in Geography and Mathematics, Oisín wants to calculate the distance traveled by an oceanographic research vessel sailing along the…
For a school project in Geography and Mathematics, Oisín wants to calculate the distance traveled by an oceanographic research vessel sailing along the $60^\circ\text{ N}$ parallel between two weather buoys located at longitudes $10^\circ\text{ W}$ and $40^\circ\text{ W}$. Oisín performs the following steps in his notebook:
$$\Delta = 40^\circ - 10^\circ = 30^\circ = 30\times\frac{\pi}{180} = \frac{\pi}{6}\text{ rad}$$ $$\text{Distance} = 6371\times\frac{\pi}{6} \approx 3335.82\text{ km}$$- a) Explain with reasons the conceptual mistake Oisín made and along which circumference of the Earth his numerical result would be valid.
- b) Correct his solution by indicating the complete formula he should have used, and calculate the actual distance traveled by the vessel (take $R = 6371\text{ km}$ and use the exact value $\cos 60^\circ = 0.5$).
Two cities, A and B, are on the same parallel at a latitude of $45^\circ \text{ N}$. City A has a longitude of $20^\circ \text{ E}$ and city B a longitude of…
Two cities, A and B, are on the same parallel at a latitude of $45^\circ \text{ N}$. City A has a longitude of $20^\circ \text{ E}$ and city B a longitude of $70^\circ \text{ E}$.
Given that the radius of the Earth is approximately $6371 \text{ km}$:
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Calculate the radius of the $45^\circ \text{ N}$ parallel (Remember that the radius of a parallel at a latitude $\alpha$ is calculated as $r = R \times \cos \alpha$; for $45^\circ$ use the exact value $\cos 45^\circ = \frac{\sqrt{2}}{2}$).
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Determine the distance in kilometres separating both cities along the arc of the parallel.
Along a meridian: comparing routes
A meridian is a line of constant longitude, running from pole to pole. It is part of a circle with the same radius $R$ as the Earth, so the radius of the circle does not shrink: the distance between two points on the same meridian that differ in latitude by $\theta$ is the fraction $\dfrac{\theta}{360}$ of $2\pi R$.
On the east coast of Ireland, Dublin is located on the Greenwich meridian ($0^\circ$) at a latitude of $40^\circ\text{ N}$. Off the west coast, the western tip of Achill…
On the east coast of Ireland, Dublin is located on the Greenwich meridian ($0^\circ$) at a latitude of $40^\circ\text{ N}$. Off the west coast, the western tip of Achill Island lies virtually on that same parallel ($40^\circ\text{ N}$) at a longitude of $3.8^\circ\text{ E}$. At the same time, a Naval Service patrol boat in the Aran Islands area carries out surveillance between two points that also differ by $3.8^\circ$ in longitude, but are located along the $28^\circ\text{ N}$ parallel. Consider the radius of the Earth $R = 6371\text{ km}$.
- a) Model the general formula that allows calculating the distance along a parallel in kilometers from the Earth's radius $R$, the latitude $\alpha$, and the difference in longitude in degrees $\Delta^\circ$.
- b) Calculate the distance in kilometers of the route Dublin–Achill Island sailing along the $40^\circ\text{ N}$ parallel (use $\cos 40^\circ \approx 0.7660$).
- c) Calculate the distance traveled by the patrol boat near the Aran Islands sailing along the $28^\circ\text{ N}$ parallel (use $\cos 28^\circ \approx 0.8829$).
- d) Even though both routes span the same longitude angle ($3.8^\circ$), the distances over the sea are not equal. Justify this discrepancy mathematically from the behavior of the function $\cos\alpha$ between $0^\circ$ and $90^\circ$, and analyze what implications this has for maritime and aerial navigation.
Two polar research bases are located on the $75^\circ\text{ N}$ parallel at diametrically opposite points: Base Alpha at longitude $0^\circ$ (Greenwich meridian) and…
Two polar research bases are located on the $75^\circ\text{ N}$ parallel at diametrically opposite points: Base Alpha at longitude $0^\circ$ (Greenwich meridian) and Base Beta at longitude $180^\circ$. To transport supplies between them, an aviation team must choose between two possible flight routes over the Earth's surface (take $R = 6371\text{ km}$):
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Route 1 (Parallel route): Fly along the circumference of the $75^\circ\text{ N}$ parallel from Alpha to Beta (half a parallel).
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Route 2 (Polar route): Fly north along the Greenwich meridian to the North Pole ($90^\circ\text{ N}$) and continue descending along the $180^\circ$ meridian until reaching Beta.
- a) Mathematically model the length of Route 1 using the radius of the parallel $r = R\times\cos(75^\circ)$ and calculate its distance in kilometers (use $\cos 75^\circ \approx 0.2588$).
- b) Mathematically model the length of Route 2 taking into account that meridians form great circles of radius $R$. Deduce how many degrees of meridian arc are traveled in total and calculate the distance in kilometers.
- c) Compare both results: which route is shorter and how many kilometers are saved? Using this calculation, justify why many intercontinental commercial flight routes in the Northern Hemisphere fly close to the pole instead of maintaining a constant latitude.
Directions and angles: reading a wind bearing
Weather forecasts such as those from Met Éireann give the wind direction as the bearing it blows from. Here we use the bearing $\alpha$ of the direction the wind is blowing towards: an angle measured clockwise from north, between $0^\circ$ and $360^\circ$, and a speed $v$. The east component of the wind is $v\sin\alpha$ and the north component is $v\cos\alpha$. A forecast of a wind "from" $35^\circ$ means it blows towards $215^\circ$ (add or subtract $180^\circ$). To find the signs of $\sin\alpha$ and $\cos\alpha$, locate $\alpha$ on the unit circle and reduce it to the first quadrant.
Topic Challenge. How does the wind affect the flight of a rescue drone? You will analyze the conditions of your environment using official Met Éireann data. Stage 1. Go…
Topic Challenge. How does the wind affect the flight of a rescue drone? You will analyze the conditions of your environment using official Met Éireann data.
Stage 1. Go to the Met Éireann website and collect the heading $r$ (measured clockwise from north) and wind speed ($v$ in km/h) from your nearest station at three times of day. Convert each heading into the directed angle $\alpha = 90^\circ - r$ (add $360^\circ$ if it is negative), so that $\alpha \in [0^\circ, 360^\circ)$ is measured counterclockwise from east.
Stage 2. In GeoGebra or in your notebook, place each angle on the unit circle. Reduce each angle to the first quadrant, deduce the signs of $\sin\alpha$ and $\cos\alpha$ according to the quadrant, and numerically check the fundamental identity $\sin^2\alpha + \cos^2\alpha = 1$ to two decimal places (e.g., $0.99 \approx 1$).
Stage 3. Trigonometrically decompose the wind speed into its east-west ($v_x = v \times \cos\alpha$) and north-south ($v_y = v \times \sin\alpha$) axes.
Stage 4. Submit a brief report (one single-sided page) with the data table, the unit circle representations, and a reasoned conclusion on which of the three times offers the least lateral drift for a fixed north-south trajectory.
Assessment criteria:
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Rigor in the reduction to the first quadrant and geometric justification of the signs.
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Correct application of the fundamental identity and trigonometric ratios with real data.
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Clarity of the report, precision in calculations, and relevance of the final recommendation.
Back to the start: at $53^\circ$ N the Earth's spin carries you round at about $1{,}000$ km/h, about $\cos 53^\circ \approx 60\%$ of the speed at the equator, because your parallel is smaller.