Formulas

L2MAEL

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Please note that this is only a proportion of L2 Trigonometry

Rule	Formula
Sine rule (sides)	$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
Sine rule (angles)	$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
Cosine rule (sides)	$a=\sqrt{b^2+c^2-2(b)(c)\cos A}$
Cosine rule (angles)	$A={\cos }^{-1}\frac{b^2+c^2-a^2}{2(b)(c)}$
Area of triangle	$\frac{1}{2}bc\sin A$
Length of arc	$2\pi (r)(\frac{\theta }{360})$
Area of a sector	$\pi (r{)}^2(\frac{\theta }{360})$
Capital abc’s such as $A$, stand for the angles, while the lowercases stand for the sides.
Note that, all $A$, or $a$ used in this note are angles/sides that are unknown. $B,b$, and $C,c$ are random depending on context.

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1. Sine Rules

1-1: Finding Sides

We use the [Sine rule], when there are 2+ opposites involved:

‘Opposites’ are opposites just like in trig’s SOH CAH TOA.

The opposite of $58^o$ = $x$, while the opposite of $61^o$ = $12cm$ (they are marked using colors for easier understanding).

When applying equations, we only need to use two sides with opposing angles:

$\frac{a}{\sin A}=\frac{b}{\sin B}$

$\frac{x}{\sin 58}=\frac{12}{\sin 61}$

Rearrange:

$x=\frac{12\sin 58}{\sin 61}$

$x=11.64cm\text{ (2dp)}$

1-2: Finding Angles

We use the same [Sine rule], as there are 2+ opposites involved:

Solve:

$\frac{\sin x}{12}=\frac{\sin 58}{11.64}$

Rearrange:

$\sin x=\frac{12\sin 58}{11.64}$

$x={\sin }^{-1}(\frac{12\sin 58}{11.64})$

$x=61\text{ (rounded)}$

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2. Cosine Rules

2-1: Finding Sides

We use the [Cosine rule], when there are no opposites involved:

As you can see from the image above, there are only 1 opposite, meaning we cannot use the sine rule.

So, we have to use the side cosine rule, with the formula being: $a=\sqrt{b^2+c^2-2(b)(c)\cos A}$

$a$, is the side that we’re looking for, and in this case, $a$ would be $x$ on the diagram. $b$ and $c$ can be any adjacent sides within the angle you were given, and in the diagram above, it is $9cm$ and $10cm$, marked by a pink highlighter.

And lastly, $A$, is the angle you were given.

Substituting values:

$x=\sqrt{9^2+10^2-2(9)(10)\cos 27}$

Solving (use calculator):

$4.54cm\text{ (2dp)}$

2-2: Finding Angles

We use the same [Cosine rule], but as there are 2+ opposites involved, we use the angle cosine rule:

$A={\cos }^{-1}(\frac{b^2+c^2-a^2}{2(b)(c)})$

Solve:

$x={\cos }^{-1}(\frac{9^2+10^2-4.54^2}{2(9)(10)})$

$x=27\text{ (rounded)}$

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3. Area of Triangles

When finding the area of triangles, we use the triangle area formula:

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

Before the sin rules and other stuffs, you would be doing this:

..calculating the height of the right angled triangle using $\sin$, multiplying it by the base, then halfing it. ($\frac{1}{2}bh$)

However, this might be the easier way:

$\frac{1}{2}(8.2)(10.5)\sin 47$

$31.48c{m}^2$

..with just one line, you can solve it. So I think this may be the easier way in most cases.

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4. Length of arcs

First of all, this is an arc:

When finding the length of an arc, we use the formula:

$2\pi (r)(\frac{\theta }{360})$

..where $r$ is the radius, and where $\theta$ is the inner angle of the sector (in the example, it’s $72^o$)

Use formula:

$2\pi (r)(\frac{\theta }{360})$

$r$ = $4.4$

$\theta$ = $104$

Substitute:

$2\pi (4.4)\left(\frac{104}{360}\right)$

$7.99m$

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5. Area of a Sector

This is a sector:

The whole thing. AKA the proportion of the circle, or a pizza slice.

To find the area of a sector, we use the circle area formula, but multiplied by $\frac{\theta }{360}$.

$\theta$ is the inner angle of the sector, and in that example it’s $72^o$.

Formula:

$\pi (r{)}^2(\frac{\theta }{360})$

$r$ = $632$

$\theta$ = $135$

Substitute:

$\pi (632{)}^2(\frac{135}{360})$

$369,479.99m{m}^2$