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FM Trigonometry

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.
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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.
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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).

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Example

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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.

Picture 2

Example

Picture 3

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