A+ » VCE » Further Maths U3 & 4 Master Notes » OA3 Geometry and Measurement » FM Geometry and Measurement

# FM Geometry and Measurement

## 2.9 Introduction to Time Zones

Note: in Further Maths, it is expected that you express time in the form hrs:mins hours e.g. 1:30 hours.

### Time Zones

• As the sun orbits around the world, different parts of the world will experience a different “part of the day”. While some are in broad daylight, others are in complete darkness. We want the time of day to tell us, regardless of where we are in the world, how long we have left before the sun rises or sets. To allow this, time zones are created. Times zones are regions which share a common time. Different time zones are set to different times to give a depiction of when the sun is expected to rise/set in that location.
• Time zones separated longitude. In Further Maths, we will approximate that each time zone is separated by 15o of longitude and each subsequent time zone differs by 1 hour.
• Time zones are set relative to Greenwich Mean Time (GMT), the time zone corresponding to the Greenwich meridian (0o). Locations east of the meridian are ahead of GMT, while locations west of the meridian are behind of GMT.

### Time in a different Time Zone

Read More »2.9 Introduction to Time Zones

## 2.8 Shortest Distance between Points on a Parallel

Note: if you cannot remember what parallels of longitude are, revise notes for 2.6 Modelling the World.

Note: if you cannot remember how to find the shortest distance between points on great and small circles, revise notes for 2.5 Distance Between Points on a Sphere.

### Shortest Distance between Points on the Equator

• Locations of equal latitude exist on the same parallel of longitude. By taking the shorter section of that parallel with the locations as the endpoints, we can extract an arc representing the shortest path between the two locations. The length of that path can be found using the longitudinal coordinates of the locations.
• The plane of the equator intersects the centre of the world and so it is a great circle.
• We can modify the equation for arc length (from notes 2.1 Circles and Arcs) for this situation. There are 3 situations with different formulas:
• If both locations are given in the same units (oE or oW) or one or both lie on the Greenwich meridian, the formula is:
Read More »2.8 Shortest Distance between Points on a Parallel

## 2.7 Shortest Distance to the Poles or Equator

Note: if you cannot remember what a meridian is, revise notes for 2.6 Modelling the World.

### Finding the Shortest Distance using the Meridian

• The shortest path between any point on the surface of the earth to the north or south pole, or the equator is along the meridian intersecting that point. By taking the section of the meridian between the relevant pole or equator and the point being solved for, we find an arc whose length is the shortest distance between the pole or equator and the position. The angle of that arc can be found using the latitude of that location.
• We can modify our formula for arc length (from 2.1 Circles and Arcs) to be specific for this situation:

s_{\text {north pole }}=\frac{\pi 6400\left(90-l_{N}\right)}{180}=\frac{\pi 640\left(90-l_{N}\right)}{18}

Read More »2.7 Shortest Distance to the Poles or Equator

## 2.6 Modelling the World

### The World as a Sphere

• The world can be approximated to be a sphere of radius 6400km.

### Meridians

• Meridians of longitude are circles drawn around the world with the same radius as the world and intercepting the north and south poles.

Example

In the above spherical representation of the world, a meridian of longitude is shown in red.

### Parallels

Read More »2.6 Modelling the World

## 2.5 Distance Between Points on a Sphere

Note: if you cannot remember how to utilise trigonometry in a circle, revise notes for 2.4 Applications of Trigonometry and Pythagoras Theorem.

Note: if you cannot remember how to calculate arc length, revise notes for 2.1 Circles and Arcs.

### Great Circles

• A circle drawn on the surface of a sphere whose radius is equal to that of the sphere is known as a great circle.

### Distance Between Points on a Great Circle

• For two points on a great circle, the path corresponding to the shortest distance between those two points is an arc. We can use our understanding of arc length (from notes 2.1 Circles and Arcs) to solve for the shortest distance.
Read More »2.5 Distance Between Points on a Sphere

## 2.4 Applications of Trigonometry and Pythagoras Theorem

Note: if you cannot remember the trigonometric identities, revise notes for 2.3 Solving Triangles using Trigonometry.

### Pythagoras’ Theorem

• Pythagoras’ theorem governs the relationship between the lengths of the sides of a right-angled triangle:

a^{2}+b^{2}=c^{2}

Where c is the length of the hypotenuse and a and b are the lengths of the other sides (note that it doesn’t matter which side is chosen to be a and which is chosen to be b).

Example

We wish to find the length of the unknown side in the right-angled triangle above. As the unknown side is not the hypotenuse, we will first have to rearrange our formula to make either a or b the subject. In this case, we will choose a:

a^{2}+b^{2} =c^{2}

a^{2} =c^{2}-b^{2}

a =\sqrt{c^{2}-b^{2}}

Now, we substitute in the known values:

a=\sqrt{15^{2}-8^{2}}=12.69

### Trigonometry in a Circle

Read More »2.4 Applications of Trigonometry and Pythagoras Theorem

## 2.3 Solving Triangles using Trigonometry

### Using Sine

• The trigonometric identity for sine tells us that for a right-angled triangle:

\sin (\theta)=\frac{O}{H}

Where O is the length of the edge opposite the angle θ and H is the length of the hypotenuse.

Example

We wish to find the unknown angle, θ, in the above right-angled triangle. As we have the lengths of the hypotenuse and opposite side, we can use sine to do this:

\sin (\theta) =\frac{O}{H}=\frac{5}{7}

\theta =\sin ^{-1}\left(\frac{5}{7}\right)=45.58^{\circ}

### Using Cosine

Read More »2.3 Solving Triangles using Trigonometry

## 2.2 Area of Sectors and Segments

Note: if you cannot remember how to calculate values of a circle, review notes for 2.1 Circles and Arcs.

### Area of a Sector

• The area of a circle sector can be found by multiplying the formula for the area of a regular circle by the fraction of the circle which the sector encompasses. The angle is used for this fraction:

A=\frac{\pi r^{2} \theta}{360}

Where A is the area of the sector, r is the radius of the sector and θ is the angle of the sector.

Example

Above is a diagram of a machine component which is manufactured by cutting a 45o sector out of circle of radius 15mm. The company making these parts wishes to know the area of material which is wasted (cut out) per component. To do this we will use our area formula:

A=\frac{\pi r^{2} \theta}{360}=\frac{\pi 15^{2} 45}{360}=88.36 m m^{2}

### Chord

Read More »2.2 Area of Sectors and Segments

## 2.1 Circles and Arcs

### Circle

• The set of all points in the plane that are a fixed distance (the radius, r) from a fixed point (the centre, O).
• The angle at the centre of a circle is 360°.

### Diameter (D)

• A chord passing through the centre.
• D = 2r, twice the length of the radius.

### Circumference of a Circle

• The circumference of a circle, AKA the perimeter, is given by:

c=2 \pi r

Read More »2.1 Circles and Arcs

## 1.7 Three-Figure Bearings

### Three-Figure Bearings

• Three-Figure bearings describe a direction in a 2D plane in terms of the angle, measured in a clockwise direction, a line drawn in that direction makes with the positive y-axis (north).
• By convention, the origin is chosen to be the point (x, y)=(0,0) on a graph.

Example

The compass direction north-east (NE) corresponds to a bearing of 315o.