# Sine Rule - Formula, Revision and Worksheets

GCSE 6 - 7AQAEdexcelOCRWJECAQA 2022Edexcel 2022OCR 2022WJEC 2022

## The Sine Rule

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

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

Level 6-7 GCSE    ## The Sine Rule Formula

Looking at the triangle below, 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 the sine rule to find missing angles and missing sides. Level 6-7 GCSE    ## Example 1: Sine rule to find a length

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

[2 marks] First we need to match up the letters in the formula with the sides we want, here:

$a=x$, $A=21\degree$, $b = 23$ and $B = 35\degree$ Next we’re ready to substitute the values into the formula. Doing so gives us:

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

Multiplying both sides by $\sin(21°)$:

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

Putting this into a calculator we get:

$x=14.37029543...$

$x=14.4$  ($3$ sf)

As in previous topics, there is no need to evaluate the sine functions until the final step.

Level 6-7GCSE    ## Example 2: Sine rule to find an angle

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

[2 marks] As we have been asked to find a missing angle, we can use another version of the  sine rule:

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

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

$x= \sin^{-1}\bigg(\dfrac{43\sin(33°)}{25}\bigg)$

$x=69.5175049...°$

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 obtuse angle, simply subtract the acute angle from $180$:

$180-69.5175049=110.4824951$

$x=110\degree$  ($2$ sf)

Level 6-7GCSE    ## 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: $\dfrac{x}{\sin(46\degree)}=\dfrac{10.5}{\sin(94\degree)}$

Solving for $x$  we get:

$x=\dfrac{10.5}{\sin(94\degree)}\times\sin(46\degree)=7.571511726...$

$x=7.57$ ($3$ sf).

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 $\sin(30\degree)$:

$x=\dfrac{5}{\sin(80\degree)}\times\sin(30\degree)=2.538566...$

$x=2.54$ cm ($3$ sf).

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:

$\sin(x)=\dfrac{12\times\sin(15\degree)}{7}=0.4436897916$

Taking the inverse sine of both sides:

$x=\sin^{-1}(0.4436897916)=26.33954244\degree$

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 $180$:

$180\degree-26.33954244\degree =153.6604576$

$=154\degree$ ($3$ sf).

Instead of typing the full number into the calculator for each step of the calculation, you can use the ANS button to save time.

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 that:

$\sin(x)=6.5 \times \dfrac{\sin(52\degree)}{12} =0.4268391582$

Taking the inverse sine of both sides and keeping the answer from the previous step on our calculator, we get:

$x=\sin^{-1}(ANS)=25.26713177$

$x=25.3 \degree$ ($3$ sf).

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:

$x=\dfrac{6}{\sin(68\degree)}\times\sin(35\degree)=3.711732685...$

$x=3.71$ cm ($3$ sf).

## Related Topics

Level 6-7GCSEKS3

#### Rearranging Formulae

Level 4-5Level 6-7GCSEKS3

## Worksheet and Example Questions

### (NEW) Sine Rule Exam Style Questions -MME

Level 6-7 GCSENewOfficial MME

### (NEW) Sine and Cosine rule mixed Exam Style Questions -MME

Level 6-7 Level 8-9 GCSENewOfficial MME

Level 6-7 GCSE

Level 6-7 GCSE

Level 6-7 GCSE

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