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# ACMGM034

## 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

## 1.6 Solving Triangles in 2 and 3 dimensions

Note: if you cannot remember the trigonometric identities, revise notes for 1.5 Methods for Solving Triangles.

### Solving for Area without Height

• In some cases, triangles will be given with the edge lengths rather than the height. In this case our regular formula for the area of a triangle (A=\frac{1}{2} b h) needs some reworking. If we draw a line down the middle of a triangle, we can effectively split it into two right-angled triangles from which we can use our trigonometric identities to solve for the height. Using our formula for sine, we find:

\sin (A) =\frac{h}{c}

h =c \sin (\theta)

Consequently:

A=\frac{1}{2} b c \sin (\theta)

• This formula works for non-right angled triangles as well.

Note: trigonometry is explored in more detail in notes 2.3 Solving Triangles using Trigonometry and 2.4 Applications of Trigonometry and Pythagoras Theorem.

Read More »1.6 Solving Triangles in 2 and 3 dimensions

## 1.5 Methods for Solving Triangles

### The 180o Rule

• For any given triangle, the sum of the inside angles is equal to 180o:

A+B+C=180^{\circ}

Where A, B and C are the inside angles of a triangle.

Example

We wish to find the unknown angle, X, in the above triangle. As the other angles are known, we can use the 180o rule. First, we substitute in the known values:

X+35+45=180

Now we solve for the unknown angle:

X+80 &=180

X &=100^{\circ}

### The Sine Rule

Read More »1.5 Methods for Solving Triangles