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Year 9AC9M9M03

The Basics

Solve spatial problems, applying angle properties, scale, similarity, Pythagoras' theorem and trigonometry in right-angled triangles.

Three ratios. Learn them, and learn the special angles.

Method

Sine
sin⁡θ=oppositehypotenuse=yr\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{r}
Opposite over hypotenuse.
Cosine
cos⁡θ=adjacenthypotenuse=xr\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{r}
Adjacent over hypotenuse.
Tangent
tan⁡θ=oppositeadjacent=yx\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{y}{x}
Opposite over adjacent.
Special angles — LEARN these
0∘30∘45∘60∘90∘sin⁡θ01222321cos⁡θ13222120tan⁡θ03313−\begin{array}{c|ccccc} & 0^\circ & 30^\circ & 45^\circ & 60^\circ & 90^\circ \\ \hline \sin\theta & 0 & \frac{1}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{3}}{2} & 1 \\ \cos\theta & 1 & \frac{\sqrt{3}}{2} & \frac{\sqrt{2}}{2} & \frac{1}{2} & 0 \\ \tan\theta & 0 & \frac{\sqrt{3}}{3} & 1 & \sqrt{3} & - \end{array}
The calculator gives excellent hints if you forget.

Worked example

e.g. in a right angled triangle, opp=3, adj=4, hyp=5\text{in a right angled triangle, } opp = 3,\ adj = 4,\ hyp = 5
Right angled triangle with sides 3, 4 and 5θopp = 3hyp = 5adj = 4Sides labelled from the angle θ
  1. Label the sides from the angleopp=3, adj=4, hyp=5opp = 3,\ adj = 4,\ hyp = 5
    Sides labelled relative to θθopp = 3hyp = 5adj = 4
  2. Sine is opposite over hypotenusesin⁡θ=35\sin\theta = \tfrac{3}{5}
  3. Cosine is adjacent over hypotenusecos⁡θ=45\cos\theta = \tfrac{4}{5}
  4. Tangent is opposite over adjacenttan⁡θ=34\tan\theta = \tfrac{3}{4}

Practice

Determine the size of the following identities (leave your answer in surds):

1
sin⁡30∘\sin 30^\circ
Answer12\tfrac{1}{2}
2
cos⁡45∘\cos 45^\circ
Answer22\tfrac{\sqrt{2}}{2}
3
tan⁡0∘\tan 0^\circ
Answer00
4
sin⁡60∘\sin 60^\circ
Answer32\tfrac{\sqrt{3}}{2}
5
cos⁡30∘\cos 30^\circ
Answer32\tfrac{\sqrt{3}}{2}
6
tan⁡45∘\tan 45^\circ
Answer11
7
sin⁡90∘\sin 90^\circ
Answer11
8
cos⁡60∘\cos 60^\circ
Answer12\tfrac{1}{2}
9
tan⁡60∘\tan 60^\circ
Answer3\sqrt{3}
10
tan⁡30∘\tan 30^\circ
Answer33\tfrac{\sqrt{3}}{3}
Next step
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