The Sine Rule Worksheets | Questions and Revision | MME

# Sine Rule – Formula, Revision and worksheets.

Level 6 Level 7

## What you need to know

When we first learn the sine functions, we learn how to use them to find missing side-lengths & angles in right-angled triangles. The sine rule is an equation that help us find missing side-lengths and angles in any triangle.

Make sure you are happy with the following topics before continuing:

With this triangle in mind, the sine rule is

$\dfrac{\textcolor{limegreen}{a}}{\sin \textcolor{limegreen}{A}}=\dfrac{\textcolor{blue}{b}}{\sin \textcolor{blue}{B}}=\dfrac{\textcolor{red}{c}}{\sin \textcolor{red}{C}}$

In this topic, we’ll go through examples of how to use both of these rules.

## Example: Sine rule to find a length

Use the sine rule to find the side-length marked $x$ to $3$sf.

[2 marks]

First we need to match up the letters in the formula with the sides we want. ,

$a=x$, $A=21\degree$, $b = 23$ and $B = 35\degree$

Now, we’re ready to substitute the values into the formula. Doing so, we get

$\dfrac{x}{\sin(21)}=\dfrac{23}{\sin(35)}$

Multiply both side by $\sin(21)$ to get

$x=\dfrac{23}{\sin(35)}\times\sin(21)$

Putting this into a calculator, we get:

$x=14.4$ to 3sf.

Like before, no need to evaluate the sine functions until the final step.

Note: we only ever use 2 parts of the sine rule at a time.

## Example: Sine rule to find an angle

Use the sine rule to find the obtuse angle marked $x$ to $2$sf.

[2 marks]

So, the flipped-over sine rule looks like

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$

You don’t have to do this, but it makes life easier.

$A=x$, $a=43$, $b=25$ and $B=33\degree$.

Subbing these values into the formula, we get

$\dfrac{\sin x}{43}=\dfrac{\sin(33)}{25}$

Multiply both sides by 43 to get

$\sin x=\dfrac{43\sin(33)}{25}$

Then, taking $\sin^{-1}$ of both sides, we get

$x=70\degree$ to 2sf.

However, the question asked for an obtuse angle, but we got an acute answer – why?

It’s because we can draw two different (but both correct) triangles using the information we were given at the start.

This is the ambiguous case of the sine rule, and it occurs when you have 2 sides and an angle that doesn’t lie between them.

To find the obtuseangle, simply subtract the acute angle from 180,

$180-70=110\degree$

### Example Questions

First, we need to find the angle opposite to the missing side as it is not given in the question. Using all the angles in a triangle add to 180 degrees we get that,

$A=180\degree-40\degree-94\degree=46\degree$

Now we have enough information to properly label the triangle and substitute values into the sine rule equation;

$\dfrac{x}{\sin(46\degree)}=\dfrac{10.5}{\sin(94\degree)}$

Multiplying both sides, and putting it into a calculator, we get

$x=\dfrac{10.5}{\sin(94\degree)}\times\sin(46\degree)=7.57\text{ (3 s.f.)}$

Here we are able to use the sine rule straightaway,

$\dfrac{x}{\sin(30\degree)}=\dfrac{5}{\sin(80\degree)}$

Multiplying both sides of the equation by $\sin(30\degree)$, we find,

$x=\dfrac{5}{\sin(80\degree)}\times\sin(30\degree)=2.54 \text{ cm (3 s.f.)}$

Here we are able to use the sine rule straightaway,

$\dfrac{\sin(x\degree)}{12}=\dfrac{\sin(15\degree)}{7}$

Multiplying both sides of the equation by 12 we find,

$\sin(x)=\dfrac{12\times\sin(15\degree)}{7}\approx0.4437$

Taking the inverse sine of both sides,

$x=\sin^{-1}(0.4437)=26.34\degree$

However considering the diagram, the angle is clearly obtuse (greater than 90 degrees). This is the ambiguous case of the sine rule, and it occurs when you have 2 sides and an angle that doesn’t lie between them. To find the obtuse angle, simply subtract the acute angle from 180,

$180\degree-26.34\degree =154\degree \text{(3 s.f.)}$

We are able to use the sine rule straightaway,

$\dfrac{\sin(x\degree)}{6.5}=\dfrac{\sin(52\degree)}{12}$

Multiplying both sides of the equation by 6.5 we find,

$\sin(x)=6.5 \times \dfrac{\sin(52\degree)}{12}$

Taking the inverse sine of both sides,

$x=\sin^{-1}(0.4268)=25.3 \degree \text{(3 s.f.)}$

Applying the sine rule,

$\dfrac{x}{\sin(35\degree)}=\dfrac{6}{\sin(68\degree)}$

Multiplying both sides of the equation by $\sin(35\degree)$, we find,

$x=\dfrac{6}{\sin(68\degree)}\times\sin(35\degree)=3.71 \text{ cm (3 s.f.)}$

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