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Math 8 · Mathematics
Geometric solids · Geometric solids

Latitude and Longitude

Before you start, make sure you know: Circumference and circle (Math 6)
Before you start, make sure you know: Solids of Revolution (Math 7)

You open the maps app on your phone in Chicago and it shows two numbers: about $41.9^\circ$ N and $87.6^\circ$ W. What do they mean, and how can two numbers pin down any spot on a whole planet? On a flat map you use $(x, y)$. On the Earth, the "sheet of paper" is a sphere, so the coordinates are angles.

In the problems on this page, use an Earth radius of $R \approx 6{,}371$ mi.

Meridians and Parallels

Picture the Earth as a sphere spinning around an axis that goes through the North Pole and the South Pole. Two families of circles help us find our way:

  • Meridians are half-circles that run from pole to pole. The reference meridian passes through Greenwich (near London). It is the Prime Meridian, $0^\circ$.

  • Parallels are circles perpendicular to the axis. The biggest one is the Equator ($0^\circ$), which splits the Earth into the Northern and Southern Hemispheres.

Drag the sliders or pick a city and watch the lines move:

The Equator and every meridian are great circles: their center is the center of the Earth, so they are as big as a circle on a sphere can be. Every other parallel is a smaller circle.

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Imagine you are a beginner cartographer. On your first map, you must locate the most important reference lines of the Earth. Look at the drawing and match each letter…

Imagine you are a beginner cartographer. On your first map, you must locate the most important reference lines of the Earth. Look at the drawing and match each letter with its corresponding name:

  1. Equator.

  2. Greenwich Meridian.

  3. North Pole.

  4. Any parallel.

A B C D
Solve
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Two ships, Alfa and Beta, are sailing in the Atlantic Ocean. Their coordinates are: Ship Alfa: $20^\circ$ N, $30^\circ$ W Ship Beta: $20^\circ$ N, $60^\circ$ W Answer…

Two ships, Alfa and Beta, are sailing in the Atlantic Ocean. Their coordinates are:

  • Ship Alfa: $20^\circ$ N, $30^\circ$ W

  • Ship Beta: $20^\circ$ N, $60^\circ$ W

Answer with reasoning:

  1. a) Are they on the same parallel or on the same meridian?
  2. b) Which of the two ships is closer to the Greenwich Meridian?
Solve

Latitude and Longitude

A globe showing the Equator, the Prime Meridian and the angles that give the coordinates of Chicago. lat NS Equator: latitude 0° Prime Meridian: 0° Chicago ≈ 41.9° N, 87.6° W Latitude first,then longitude.

To decide the hemisphere, ask two questions. North or south of the Equator? East or west of Greenwich? For example, $12^\circ$ S, $45^\circ$ W is south of the Equator and west of Greenwich.

Reading a flat map

On a flat map (a planisphere), latitude is the vertical direction and longitude is the horizontal direction. Here the lines are $10^\circ$ apart.

A flat map grid with a point Z marked, used to read its latitude and longitude. Z 20° N10° N0°10° S20° S 30° W10° W0°10° E30° E

Degrees and minutes

Angles can be split into smaller units: $1^\circ = 60'$ (60 minutes). So $3^\circ 41'$ means 3 degrees and 41 minutes. To subtract, borrow a degree: $180^\circ - 3^\circ 41' = 179^\circ 60' - 3^\circ 41' = 176^\circ 19'$.

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State in which hemisphere (North or South for latitude, East or West for longitude) the following geographical points are located: Point A: $34^\circ$ N, $120^\circ$ E…

State in which hemisphere (North or South for latitude, East or West for longitude) the following geographical points are located:

  • Point A: $34^\circ$ N, $120^\circ$ E

  • Point B: $12^\circ$ S, $45^\circ$ W

  • Point C: $89^\circ$ N, $5^\circ$ W

  • Point D: $40^\circ$ S, $170^\circ$ E

Solve
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An explorer has mistakenly written down the following coordinates in a diary. Explain why these coordinates are geographically impossible: a) Latitude: $105^\circ$ S,…

An explorer has mistakenly written down the following coordinates in a diary. Explain why these coordinates are geographically impossible:

  1. a) Latitude: $105^\circ$ S, Longitude: $20^\circ$ E.
  2. b) Latitude: $45^\circ$ N, Longitude: $210^\circ$ W.
Solve
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Look at the following flat representation (planisphere) of an ocean area. Identify the coordinates (Latitude and Longitude) of point $P$ marked with a star.

Look at the following flat representation (planisphere) of an ocean area. Identify the coordinates (Latitude and Longitude) of point $P$ marked with a star.

P 20ºN 10ºN 0º 10ºS 20ºS 20ºE 30ºE 40ºE 50ºE 60ºE Equator N S E W
Solve

Antipodes

The antipode of a place is the point on the exact opposite side of the Earth, straight through the center. To find it:

  • Latitude: same number, opposite hemisphere (N becomes S).

  • Longitude: subtract from $180^\circ$ and switch E/W.

Why? Going through the center flips north to south. For longitude, the opposite meridian is half a turn ($180^\circ$) around the Earth.

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If you travel by airplane from Chicago ($40^\circ$ N, $3^\circ$ W) to a city that is exactly at its antipodes (the opposite point on the Earth's sphere): a) What would…

If you travel by airplane from Chicago ($40^\circ$ N, $3^\circ$ W) to a city that is exactly at its antipodes (the opposite point on the Earth's sphere):

  1. a) What would the latitude of that point be? (Same value but opposite direction).
  2. b) If the longitude of a point and its antipode must add up to $180^\circ$ and be in opposite hemispheres, what would the longitude of that point be?
Solve
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The antipodes of a place are said to be the diametrically opposite point on the Earth's surface. If the coordinates of your city are $40^\circ$ $25'$ N latitude and…

The antipodes of a place are said to be the diametrically opposite point on the Earth's surface. If the coordinates of your city are $40^\circ$ $25'$ N latitude and $3^\circ$ $41'$ W longitude:

  1. a) What would the latitude of your antipodes be?
  2. b) What would the longitude of your antipodes be?

(Hint: For the latitude, change the hemisphere; for the longitude, find the meridian that completes $180^\circ$ and change the E/W hemisphere).

N S Equator 3°41'W meridian Opposite meridian City 40°25'N, 3°41'W 40º25' Antipode Lat?, Lon?
Solve

Distance Along a Meridian

Two cities on the same meridian are joined by an arc of a great circle. A full trip around the Earth is $2\pi R \approx 2\pi(6{,}371) \approx 40{,}030$ mi, so each degree of arc is $40{,}030 \div 360 \approx 111.2$ mi.

A side view of the Earth with an arc of a meridian between two points and the central angle between them. A 30° NB 20° S 50° arc = 50/360of a full circle
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Two cities, Sunville and Moonville, are located on the same meridian. If Sunville has a latitude of $45^\circ$ N and Moonville has a latitude of $20^\circ$ S, how many…

Two cities, Sunville and Moonville, are located on the same meridian. If Sunville has a latitude of $45^\circ$ N and Moonville has a latitude of $20^\circ$ S, how many degrees of meridian arc separate them?

Equator 45° 20° Sunville Moonville
Solve
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A humpback whale migrates following exactly the same meridian from a feeding ground located at $65^\circ$ N latitude to a breeding ground at $15^\circ$ S. If the mean…

A humpback whale migrates following exactly the same meridian from a feeding ground located at $65^\circ$ N latitude to a breeding ground at $15^\circ$ S. If the mean radius of the Earth is considered to be $6\,371$ miles, calculate the total distance in miles that the whale travels on its outward journey.

N Pole S Pole Equator R = 6,371 miles 65° N 15° S 65° 15°
Solve
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Calculate the shortest distance between two points, P and Q, located on the Earth's surface if their coordinates are: P: $(40^\circ \text{ N}, 10^\circ \text{ O})$ Q:…

Calculate the shortest distance between two points, P and Q, located on the Earth's surface if their coordinates are:

  • P: $(40^\circ \text{ N}, 10^\circ \text{ O})$

  • Q: $(40^\circ \text{ S}, 10^\circ \text{ O})$

Consider the Earth to be a perfect sphere of radius $6371 \text{ mi}$. Express the result in miles. Does the traveler move along a meridian or a parallel? Justify your answer.

Equator N S 10° W meridian d 40° 40° R = 6,371 miles C P Q
Solve

Distance Along a Parallel

A parallel is a smaller circle. To find how far you travel along it, you first need its radius. Slice the Earth through the axis and look at the right triangle formed by the Earth's radius $R$, the parallel's radius $r$, and the latitude angle $\alpha$ at the center.

A cross-section of the Earth showing that the radius r of a parallel at latitude alpha equals R times cosine alpha. α R r r = R · cos α The parallel is a circlewith radius r.

Why cosine? In the right triangle, $r$ is the side next to angle $\alpha$ and $R$ is the hypotenuse, so $\cos\alpha = \dfrac{r}{R}$. Some values to remember: $\cos 30^\circ \approx 0.866$, $\cos 40^\circ \approx 0.766$, $\cos 45^\circ \approx 0.707$, $\cos 60^\circ = 0.5$.

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For a school project, you need to calculate the length of the $40^\circ$ N parallel of latitude. Knowing that the Earth's radius is $R \approx 6\,371$ miles, calculate:…

For a school project, you need to calculate the length of the $40^\circ$ N parallel of latitude. Knowing that the Earth's radius is $R \approx 6\,371$ miles, calculate:

  1. a) The radius of the circle formed by this parallel (use the expression $r = R \cdot \cos(\text{latitud})$).
  2. b) The total length (perimeter) of this parallel.
R r 40° N S Equator 40° parallel
Solve
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Two cities, P and Q, are located on the same meridian. City P is at $42^\circ$ N and city Q is at $15^\circ$ N. Another city, R, is located on the same parallel as Q,…

Two cities, P and Q, are located on the same meridian. City P is at $42^\circ$ N and city Q is at $15^\circ$ N. Another city, R, is located on the same parallel as Q, but at a distance of $20^\circ$ of longitude to the East.

Calculate, using $R = 6\,371$ miles:

  1. a) The distance between P and Q along the meridian.
  2. b) The distance between Q and R along the parallel.

Which of the two distances is greater?

Equator 42°N 15°N P Q d(P,Q) = ? R 20° d(Q,R) = ? R = 6371 miles
Solve
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Two cities, A and B, are located 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 located 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 $70^\circ \text{ E}$.

Knowing that the radius of the Earth is approximately $6371 \text{ mi}$:

  1. Calculate the radius of the $45^\circ \text{ N}$ parallel (Remember that the radius of a parallel at a latitude $\alpha$ is calculated as $r = R \cdot \cos \alpha$. For $45^\circ$, you can use $0,707$ as an approximation of the cosine).

  2. Determine the distance in miles that separates both cities following the arc of the parallel.

45° R = 6,371 miles r 50° A (20° E) B (70° E) 45° N parallel Equator
Solve

Working Backward and Crossing 180°

Sometimes you know the distance and need the angle. Reverse the arc formula: $\alpha = \dfrac{d\cdot 360}{2\pi R}$. On the Equator (or any great circle), each $111.2$ mi is $1^\circ$.

The short way around

Longitude difference between two places:

  • Same side of the Prime Meridian (both E or both W): subtract.

  • Opposite sides (one E, one W): add.

  • If the answer is over $180^\circ$, the short way goes across the opposite side of the globe. Use $360^\circ - \text{answer}$.

A line of longitudes from 180 degrees west to 180 degrees east, showing that the short way from 175 degrees east to 170 degrees west wraps around the edge. 180° W0°180° E A 175° EB 170° W short way: 5° + 10° = 15°

Two more tools:

  • Nautical miles. A nautical mile is the length of a meridian arc of $1'$. Since $60' = 1^\circ$, convert minutes to degrees by dividing by $60$, then use the distance per degree.

  • Speed. Time $=$ distance $\div$ speed. A knot is one nautical mile per hour, and the problem tells you how to convert knots to mph.

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Two oceanographic research ships are on the Equator. Ship A is at a longitude of $12^\circ$ E and ship B is at a distance of $2\,500$ miles to the West of ship A.…

Two oceanographic research ships are on the Equator. Ship A is at a longitude of $12^\circ$ E and ship B is at a distance of $2\,500$ miles to the West of ship A. Determine the exact longitude of ship B, knowing that the radius of the Earth is $6\,371$ miles.

Equator R = 6,371 miles 0° (Greenwich) 12° ? A (12°E) B (?) 2,500 miles North Pole ← WEST EAST →
Solve
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In navigation, the "nautical mile" is used, which is defined as the length of a meridian arc of $1'$ (one minute of a degree). a) Calculate how many miles a nautical…

In navigation, the "nautical mile" is used, which is defined as the length of a meridian arc of $1'$ (one minute of a degree).

  1. a) Calculate how many miles a nautical mile is equivalent to (Radius of the Earth = $6\,371$ miles).
  2. b) If a sailboat has sailed along a meridian from a latitude of $30^\circ$ N to $35^\circ$ N, how many nautical miles has it traveled?
Equator N Pole W R = 6371 miles 30°N 35°N Distance travelled 5° 1 Nautical Mile 1' 1 nautical mile R = 6371 miles
Solve
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A submarine travels along the seabed from a point X $(0^\circ, 175^\circ \text{ E})$ to a point Y $(0^\circ, 170^\circ \text{ O})$. 1. How many degrees of longitude has…

A submarine travels along the seabed from a point X $(0^\circ, 175^\circ \text{ E})$ to a point Y $(0^\circ, 170^\circ \text{ O})$.

  1. How many degrees of longitude has it traveled if it followed the shortest route along the Equator?

  2. If the submarine travels at a constant speed of $20 \text{ nudos}$ ($1 \text{ nudo} \approx 1,85 \text{ mph}$), how long will it take to complete the journey? (Radius of the Earth: $6371 \text{ mi}$).

  3. Was there a change of date during the journey? Explain why.

5º 10º Δλ = 15º X (175° E) Y (170° W) North Pole 180° (Antimeridian) 0° (Greenwich Meridian) Eastern Hemisphere Western Hemisphere Equator R = 6,371 miles v = 20 knots Date change when crossing
Solve

Time Zones

The Earth turns $360^\circ$ in $24$ hours, so it turns $360^\circ \div 24 = 15^\circ$ every hour. That is why the world is split into 24 time zones, each $15^\circ$ of longitude wide. On this page we use theoretical zones centered on multiples of $15^\circ$.

To find a place's zone number, divide its longitude by $15^\circ$ and round to the nearest whole number. East counts positive and west counts negative. The time difference is the difference of the zone numbers.

Consecutive time zone strips of 15 degrees of longitude each, with clock times growing toward the east. 10 a.m.11 a.m.noon 1 p.m.2 p.m.3 p.m. ← westeast → each strip: 15° of longitude = 1 hour

The Earth spins toward the east, so the sun rises first in the east. East is later, west is earlier.

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We know that the Earth takes 24 hours to complete a $360^\circ$ rotation on its axis. a) How many degrees of longitude correspond to a one-hour difference? b) If it is…

We know that the Earth takes 24 hours to complete a $360^\circ$ rotation on its axis.

  1. a) How many degrees of longitude correspond to a one-hour difference?
  2. b) If it is 12:00 noon in a city located at $0^\circ$ longitude (Greenwich), what time will it be in a city located at $30^\circ$ E?
Solve
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A group of friends in Madrid (longitude $3^\circ$ $41'$ W) wants to watch a live space rocket launch from Cape Canaveral (longitude $80^\circ$ $36'$ W). If the launch is…

A group of friends in Madrid (longitude $3^\circ$ $41'$ W) wants to watch a live space rocket launch from Cape Canaveral (longitude $80^\circ$ $36'$ W). If the launch is scheduled for 2:30 p.m. local time in Florida, and knowing that each $15^\circ$ time zone is equivalent to a theoretical one-hour difference, at what time should they tune into the live stream in Spain? Use the theoretical time zone diagram to carry out the calculation.

← WEST (W) EAST (E) → 0° Madrid 3°41' W ? h Cape Canaveral 80°36' W 2:30 p.m. Δt = ? 1 time zone = 15° = 1 h
Solve

Flights and the Date Line

When a plane flies far east or west, the clock changes during the flight. Use this plan:

  1. Add the flight time to the departure time to get the time on the departure clock when the plane lands.

  2. Add the zone difference: (destination zone $-$ departure zone) hours.

A three-step timeline of a flight: departure time, the arrival time on the departure clock, and the arrival time on the destination clock. Leave10 a.m. Wed Origin clock9 p.m. Wed Destination3 a.m. Wed +11 h−18 h zones

The plane left on Wednesday morning and arrived on Wednesday at 3 a.m. That is not a mistake. Flying east across the International Date Line (near the $180^\circ$ meridian), the calendar goes back one day. Flying west across it, the calendar goes forward one day. The zone numbers you used already take care of this, because heading east across $180^\circ$ turns the zone number from $+12$ to $-12$.

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An airplane takes off from Tokyo ($35^\circ$ N, $139^\circ$ E) at 8:00 a.m. on Monday bound for San Francisco ($37^\circ$ N, $122^\circ$ W). If the flight lasts 10…

An airplane takes off from Tokyo ($35^\circ$ N, $139^\circ$ E) at 8:00 a.m. on Monday bound for San Francisco ($37^\circ$ N, $122^\circ$ W). If the flight lasts 10 hours, what time and day will it be in San Francisco when the airplane lands? Take into account the date change when crossing the Greenwich antimeridian ($180^\circ$).

120°E 135°E 150°E 165°E 165°W 150°W 135°W 120°W UTC+8 UTC+9 UTC+10 UTC+11 UTC-11 UTC-10 UTC-9 UTC-8 Antimeridian (180°) Change: −1 day Flight: 10h Tokyo (139°E) Monday 8:00 a.m. San Francisco (122°W) Arrival: ?
Solve
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A reconnaissance plane takes off from a base located at coordinates $(30^\circ \text{ N}, 15^\circ \text{ O})$ at 8:00 a.m. local time. It flies directly East following…

A reconnaissance plane takes off from a base located at coordinates $(30^\circ \text{ N}, 15^\circ \text{ O})$ at 8:00 a.m. local time. It flies directly East following the same parallel until it reaches a city located at $60^\circ$ longitude East. The flight has a total duration of 9 hours.

At what local time of the destination city will the plane land? Take into account the change of time zones during the journey (consider theoretical time zones of $15^\circ$ each).

Greenwich 30°W 15°W 0º 15°E 30°E 45°E 60°E 75°E +1h +1h +1h +1h +1h Time zone change: +5 hours Flight: 9 hours Departure 08:00 Arrival ??:??
Solve

Back to the start: your phone's $41.9^\circ$ N, $87.6^\circ$ W is Chicago. It is $41.9^\circ$ north of the Equator and $87.6^\circ$ west of the Prime Meridian, so it is in the Northern and Western Hemispheres. Its time zone is $87.6 \div 15 \approx 5.8$, which rounds to $-6$, and that is why the clock in Chicago is six hours behind Greenwich.

Truncated Solids and Euler's Formula

The globe is a sphere, but this section also includes polyhedra problems. Some of them involve a truncated solid (a solid with its corners cut off) and need Euler's formula.

A cube checks out: $8 - 12 + 6 = 2$. To count a truncated solid, start from the original solid and ask what a cut does. If $k$ faces meet at a vertex, slicing that vertex off makes one new $k$-sided face, $k$ new vertices, and $k$ new edges. The old vertex disappears.

Area of a regular hexagon

A truncated solid has hexagon faces, so you also need the area of a regular hexagon. Draw lines from the center to the six vertices. You get $6$ equilateral triangles, and the hexagon's side $s$ equals the circumscribed radius.

A regular hexagon split into six equilateral triangles from its center, with the side s and the apothem a marked. sa side s a = (√3/2) s area = (3√3/2) s²

Why? Each equilateral triangle has base $s$ and height (the apothem) $a$. By the Pythagorean Theorem, $a = \sqrt{s^2 - (s/2)^2} = \tfrac{\sqrt3}{2}s$. Then the hexagon's area is $6\cdot\tfrac12 s a = 3sa = \tfrac{3\sqrt3}{2}s^2$. You can also use $A = \tfrac12\cdot\text{perimeter}\cdot a$.

Now put both ideas together. The next problem uses centimeters, as the exercise does.

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A truncated octahedron is constructed from a regular octahedron with an edge of $12 \text{ cm}$. To do this, each of the six vertices is cut off so that small regular…

A truncated octahedron is constructed from a regular octahedron with an edge of $12 \text{ cm}$. To do this, each of the six vertices is cut off so that small regular square-based pyramids are removed, turning each original triangular face into a regular hexagon.

  1. Determine the number of faces, vertices and edges of the resulting polyhedron and check if it satisfies Euler's formula.

  2. If the edge of the new polyhedron (that of the resulting hexagon) is $4 \text{ cm}$, calculate the total surface area of the truncated octahedron.

hexagon square a = 4 cm F = ? V = ? E = ? F + V = E + 2
Solve
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A water tank has the shape of a right regular hexagonal prism. It is known that the radius of the circumscribed circle of the hexagonal base measures $4 \text{ m}$ and…

A water tank has the shape of a right regular hexagonal prism. It is known that the radius of the circumscribed circle of the hexagonal base measures $4 \text{ m}$ and that the height of the prism is three times the apothem of the base.

Calculate the cost of waterproofing the side walls and the bottom (bottom base) if the price of the material is $12,50 \text{ \$/m}^2$. Round the result to two decimal places.

R = 4 ft ap H = 3·ap
Solve

Volumes of Frustums and Spheres

The last two problems combine round solids. Here are the volume formulas you need (they are all in terms of the radius $r$ and height $h$):

Solid Volume
Cylinder $V = \pi r^2 h$
Cone $V = \tfrac13 \pi r^2 h$
Sphere $V = \tfrac43 \pi r^3$
Hemisphere $V = \tfrac23 \pi r^3$

In these two problems the units are centimeters, and $1\ \text{cm}^3 = 1\ \text{mL}$.

A frustum of a cone is a cone with its tip cut off by a plane parallel to the base. There is no frustum formula to memorize. Instead, extend it to a full cone and subtract the small cone that was cut off. It does not matter which base is on top: a frustum standing on its small base is just this one turned upside down.

A cone with its tip cut off by a plane parallel to the base; the dashed small cone is removed and the rest is the frustum. cut-off tip frustum big cone − small cone = frustum

To find the height of the small cone, use similar triangles. The small cone and the big cone have the same shape, so their radii and heights are in the same ratio.

A sphere inside a cylinder

A sphere of radius R fitting exactly inside a cylinder of radius R and height 2R. R 2R cylinder = 2πR³ sphere = (4/3)πR³ ratio = 2/3

This is no accident. If you replace the $2$ by a letter $R$, the same steps work: cylinder $= \pi R^2(2R) = 2\pi R^3$ and sphere $= \tfrac43\pi R^3$, so the ratio is always $\tfrac23$.

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A perfume company designs a bottle in the shape of an inverted frustum of a cone topped by a hemisphere that fits perfectly on the top base. The design data is as…

A perfume company designs a bottle in the shape of an inverted frustum of a cone topped by a hemisphere that fits perfectly on the top base. The design data is as follows:

  • Height of the frustum of a cone: $8 \text{ cm}$.

  • Radius of the bottom base (where it rests): $3 \text{ cm}$.

  • Radius of the top base (where the hemisphere fits): $6 \text{ cm}$.

Calculate the total volume of the bottle and determine if it is able to hold a standard $330 \text{ ml}$ bottle of essence. (Take $\pi \approx 3,14$).

h=8 r=3 R=6
Solve
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There is a cylinder of radius $R$ and height $2R$. A sphere of radius $R$ is placed inside it, fitting tangentially with the walls and bases of the cylinder. 1. Prove…

There is a cylinder of radius $R$ and height $2R$. A sphere of radius $R$ is placed inside it, fitting tangentially with the walls and bases of the cylinder.

  1. Prove analytically the relationship between the volume of the sphere and the volume of the cylinder.

  2. If the volume of the cylinder is $54 \text{ cm}^3$, what volume of air is left free inside the cylinder once the sphere is inserted?

R R 2R
Solve

Summary

See all the exercises in this section