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In order to access this I need to be confident with:

Trigonometric functions: sine and inverse sine

Solving equations Rearranging equationsRounding to decimal places

Rounding to significant figures

Exact trigonometric values

This topic is relevant for:

Here we will learn about the **sine rule** including how to use the sine rule to find missing sides and angles in triangles.

There are also sine rule worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

The **sine rule **(or the law of sines) is a relationship between the size of an angle in a triangle and the opposing side.

We can use the sine rule to work out a missing angle or side in a triangle when we have information about an angle and the side opposite it, and another angle and the side opposite it.

This is the sine rule:

\[\frac{a}{\sin (A)}=\frac{b}{\sin (B)}=\frac{c}{\sin (C)}\]

Even though there are three relationships in a triangle between the three angles and their opposite sides, we only need to use two of them for the sine rule.

It would look something like this:

\[\frac{a}{\sin (A)}=\frac{b}{\sin (B)}\]

We can use the sine rule to find missing sides and angles when we have information about a pair of angles and their opposite sides.

1 **Label each angle (A, B, C) and each side (a, b, c) of the triangle.**

Capital letters

The opposite angle is the same letter as the opposite side.

It does not matter which sides are **must be opposite** the same letter on the side for the sine relationship to work.

2 **State the sine rule then substitute the given values into the equation.**

Now that we know which sides and angles we have, we need to substitute this information into the sine rule. We can then solve this equation to find the missing side or angle.

As these are calculated slightly differently, we can rearrange the sine rule to suit the part of the triangle we are trying to find. Here are the two versions.

To find a missing angle:

\frac{\sin (A)}{a}=\frac{\sin (B)}{b}

To find a missing side:

\frac{a}{\sin (A)}=\frac{b}{\sin (B)}

One equation is a rearrangement of the other.

Top tips:

When we are finding a **missing angle,** we put the **angles on top**

When we are finding a **missing side**, we put the **sides on top**

3 **Solve the equation.**

Once everything is substituted into the sine rule we can solve the equation to calculate the unknown side or angle.

In order to find a missing side of a triangle using the sine rule:

**Label each angle**(A, B, C) and each side(a, b, c) of the triangle.**State the sine rule then substitute the given values into the equation.****Solve the equation.**

Get your free sine rule worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREEGet your free sine rule worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREECalculate the length

**Label each angle**.(A, B, C) and each side(a, b, c) of the triangle

2 **State the sine rule then substitute the given values into the equation.**

Here we know side

\[\begin{aligned}
\frac{a}{\sin (A)}&=\frac{c}{\sin (C)}\\\\
\frac{6}{\sin (55)}&=\frac{c}{\sin (73)}
\end{aligned}
\]

3 **Solve the equation.**

\begin{array}{l}
\\\frac{6}{\sin (55)}=\frac{c}{\sin (73)}\\\\
c=\frac{6}{\sin (55)}\times\sin(73)\\\\
c=\frac{6\sin(73)}{\sin (55)}\\\\
c=7.00\quad(2d.p.)
\end{array}

Here, the length

Calculate the length

**Label each angle **** and each side **

**State the sine rule then substitute the given values into the equation.**

Here we know side

\begin{aligned}
\\\frac{a}{\sin (A)}&=\frac{c}{\sin (C)}\\
\\\frac{a}{\sin (41)}&=\frac{3.4}{\sin (27)}\\
\end{aligned}

**Solve the equation.**

\begin{array}{l}
\frac{a}{\sin (41)}=\frac{3.4}{\sin (27)}\\
\\a=\frac{3.4}{\sin (27)}\times\sin(41)\\
\\a=\frac{3.4\sin(41)}{\sin (27)}\\
\\a=4.91\quad(2d.p.)
\end{array}

Here, the length

Calculate the length

**Label each angle **

**State the sine rule then substitute the given values into the equation.**

Here we know side

\begin{aligned}
\\\frac{a}{\sin (A)}&=\frac{b}{\sin (B)}\\
\\\frac{10.8}{\sin (81)}&=\frac{b}{\sin (48)}\\
\end{aligned}

**Solve the equation.**

\begin{array}{l}
\\\frac{10.8}{\sin (81)}=\frac{b}{\sin (48)}\\
\\b=\frac{10.8}{\sin (81)}\times\sin(48)\\
\\b=\frac{10.8\sin(48)}{\sin (81)}\\
\\b=8.13\quad(2d.p.)
\end{array}

Here, the length

Calculate the size of the missing angle

**Label each angle **

**State the sine rule then substitute the given values into the equation**

Here we know the angle at

\begin{aligned}
\\\frac{\sin (B)}{b}&=\frac{\sin (C)}{c}\\
\\\frac{\sin (86)}{9.5}&=\frac{\sin (\theta)}{4.7}\\
\end{aligned}

**Solve the equation.**

\begin{array}{l}
\\\frac{\sin (86)}{9.5}=\frac{\sin (\theta)}{4.7}\\
\\\sin (\theta)=\frac{\sin (86)}{9.5}\times4.7\\
\\\sin (\theta)=0.493531688…\\
\\\theta=\sin ^{-1}(0.493531688)\\
\\\theta=29.6^{\circ}\quad(1d.p.)
\end{array}

Here, the angle

\[\theta = 29.6^{\circ} (2d.p.).\]

Calculate the size of the missing angle

**Label each angle *** *and each side** **

**State the sine rule then substitute the given values into the equation.**

Here we know the angle at

\begin{aligned}
\\\frac{\sin (B)}{b}&=\frac{\sin (C)}{c}\\
\\\frac{\sin (10)}{1.5}&=\frac{\sin (\theta)}{7.6}\\
\end{aligned}

**Solve the equation.**

\begin{array}{l}
\\\frac{\sin (10)}{1.5}=\frac{\sin (\theta)}{7.6}\\
\\\sin (\theta)=\frac{\sin (10)}{1.5}\times7.6\\
\\\sin (\theta)=0.8798174335…\\
\\\theta=\sin ^{-1}(0.8798174335)\\
\\\theta=61.6^{\circ}\quad(1d.p.)
\end{array}

Here, the length

Calculate the size of the missing angle

**Label each angle *** *and each side ** ** of the triangle.

**State the sine rule then substitute the given values into the equation.**

Here we know the angle at

\begin{aligned}
\\\frac{\sin (A)}{a}&=\frac{\sin (B)}{b}\\
\\\frac{\sin (\theta)}{5}&=\frac{\sin (60)}{5}\\
\end{aligned}

**Solve the equation.**

\begin{array}{l}
\\\frac{\sin (\theta)}{5}=\frac{\sin (60)}{5}\\
\\\sin (\theta)=\frac{\sin (60)}{5}\times5\\
\\\sin (\theta)=\frac{\sqrt{3}}{2}\\
\\\theta=\sin ^{-1}(\frac{\sqrt{3}}{2})\\
\\\theta=60^{\circ}\quad
\end{array}

Here, the length

**Incorrect labelling of the triangle**

This can** **cause a follow through error when substituting values into the equation. For example, this triangle has been incorrectly labelled with the side next to the angle.

This will have an impact on the formula for the sine rule as it will look like this, and not the correct way in example 5.

**Mixing up when to use the sine and inverse sine function**

When finding a side, use the sine function. When finding a missing angle, you need to use the inverse sine function.

**Using the sine rule to find a missing side when you do not know the value of two angles in the triangle**

In order to use the sine rule we need to have sides and angles that are opposite each other. If we are finding a missing side then we need to know two angles and one of the sides opposite, we can use this information to find the side that is opposite the other angle.

**Rounding the decimal too early**

This can lose accuracy marks. Always use as many decimal places as possible throughout the calculation, then round your solution.

**Using the incorrect set up for the sine rule formula**

It should not be too difficult to rearrange the formula once you have substituted the values, however small errors will make an impact on your solution.

**Cancelling out sine functions**

For example, when calculating the value for a in example 2, the following calculation error is made.

\begin{array}{l}
\\\frac{a}{\sin (41)}=\frac{3.4}{\sin (27)}\\
\\a=\frac{3.4}{\sin (27)}\times\sin(41)\\
\\a=\frac{3.4\sin(41)}{\sin (27)}\\
\\a=\frac{3.4\times41}{27}\\
\\a=5.16\quad(2dp)
\end{array}

This is incorrect because sine is a function so we cannot do this.

**Be aware of the ambiguous case for the sine rule**

Due to the nature of the sine function, there can be multiple solutions to a question.

Take for example

A calculator is programmed to return the first value for the angle but you should be aware that another solution can exist for an obtuse angle.

To check if the calculator has returned the solution you need, there are 3 simple checks that can be done.

Sin(θ) = 1 . This case only has one solution,θ = 90º (a right angle triangle).- The sum of the angles is greater than
180º . This would make an impossible triangle. - The angle opposite a longer side must be larger. Compare the values for the sides of the triangle
a, b, andc with their counterpart anglesA, B, andC . The larger side will always be opposite a larger angle.

1. Calculate the length BC. Write your answer to two decimal places.

6.29cm

4.95cm

6.07cm

6.11cm

\begin{array}{l} \frac{a}{\sin(A)}=\frac{c}{\sin(C)}\\\\ \frac{a}{\sin(48)}=\frac{5.5}{\sin(42)}\\\\ a=\frac{5.5}{\sin(42)} \times \sin(48)\\\\ a=6.11 \mathrm{~cm} \end{array}

2. Calculate the length PQ . Write your answer to two decimal places.

88.02mm

34.37mm

99.69mm

88.21mm

\begin{array}{l} \frac{b}{\sin(B)}=\frac{c}{\sin(C)}\\\\ \frac{55}{\sin(32)}=\frac{c}{\sin(58)}\\\\ a=\frac{55}{\sin(32)} \times \sin(58)\\\\ a=88.02 \mathrm{~mm} \end{array}

3. Calculate the length XY . Write your answer to two decimal places.

7.16cm

6.09cm

5.83cm

7.47cm

\begin{array}{l} \frac{a}{\sin(A)}=\frac{c}{\sin(C)}\\\\ \frac{6.6}{\sin(60)}=\frac{c}{\sin(53)}\\\\ a=\frac{6.6}{\sin(60)} \times \sin(53)\\\\ a=6.09\mathrm{~cm} \end{array}

4. Calculate the size of angle \theta . Write your answer to two significant figures.

66^{\circ}

70^{\circ}

0.016^{\circ}

0.93^{\circ}

\begin{array}{l} \frac{\sin(B)}{b}=\frac{\sin(C)}{c}\\\\ \frac{\sin (\theta)}{7.4}=\frac{\sin (68)}{7.3}\\\\ \sin (\theta)=\frac{\sin (68)}{7.3}\times7.4\\\\ \sin (\theta)=0.9398850033…\\\\ \theta=\sin^{-1}(0.9398850033)\\\\ \theta=70^{\circ}\quad(2 s.f.) \end{array}

5. Use trigonometry to calculate the size of angle \theta . Write your answer to two decimal places.

27.58^{\circ}

51.90^{\circ}

28.16^{\circ}

32.51^{\circ}

\begin{array}{l} \frac{\sin(A)}{a}=\frac{\sin(C)}{c}\\\\ \frac{\sin (62)}{8.2}=\frac{\sin (\theta)}{4.3}\\\\ \sin (\theta)=\frac{\sin (62)}{8.2}\times4.3\\\\ \sin (\theta)=0.4630091036…\\\\ \theta=\sin^{-1}(0.4630091036)\\\\ \theta=27.58^{\circ}\quad(2 d.p.) \end{array}

6. Calculate the size of angle \theta . Write your answer to three significant figures.

72^{\circ}

22.5^{\circ}

64.7^{\circ}

55.4^{\circ}

\begin{array}{l} \frac{\sin(A)}{a}=\frac{\sin(B)}{b}\\\\ \frac{\sin (\theta)}{10}=\frac{\sin (36)}{6.5}\\\\ \sin (\theta)=\frac{\sin (36)}{6.5}\times10\\\\ \sin (\theta)=0.9042850035…\\\\ \theta=\sin ^{-1}(0.9042850035)\\\\ \theta=64.72^{\circ}\quad(3s.f.) \end{array}

1. In triangle ABC, AC = 8.5cm, BC = 23cm and angle BAC = 80^{\circ}.

Calculate the size of the angle marked x.

Give your answer to 1 d.p.

**(3 Marks)**

Show answer

\frac{\sin(x)}{8.5}=\frac{\sin(80)}{23}

**(1)**

\begin{aligned} \sin(x)&=\frac{\sin(80)}{23} \times 8.5\\\\ \sin(x)&=0.364… \end{aligned}

**(1)**

\begin{aligned} x&=\sin^{-1}(0.364…)\\\\ x&=21.3^{\circ} \end{aligned}

**(1)**

2. Work out the size of angle BCD.

**(5 Marks)**

Show answer

\frac{BD}{\sin(77)}=\frac{14}{\sin(51)}

**(1)**

\begin{aligned} BD&=\frac{14}{\sin(51)} \times \sin(77)\\\\ BD&= 17.553\mathrm{cm} \end{aligned}

**(1)**

\frac{sin(\theta)}{17.55}=\frac{sin(82)}{30}

**(1)**

\begin{aligned} \sin(\theta)&=\frac{sin(82)}{30} \times 17.55\\\\ sin(\theta)&=0.579… \end{aligned}

**(1)**

\begin{aligned} \theta&=sin^{-1}(0.579…)\\\\ \theta&=35.4^{\circ} \end{aligned}

**(1)**

3. Calculate the length AB.

Give your answer to 1 d.p.

**(4 Marks)**

Show answer

\begin{array}{l} \text{Angle ABE}=42^{\circ} \text{ (Alternate angles are equal)}\\\\ \text{Angle AEB: }180-46-42=92^{\circ} \end{array}

**(1)**

\frac{AB}{\sin(92)}=\frac{17}{\sin(46)}

**(1)**

AB=\frac{17}{\sin(46)} \times \sin(92)

**(1)**

AB=23.6cm

**(1)**

You have now learned how to:

- know and apply the sine rule to find unknown lengths and angles

- Cosine rule
- Area of a triangle
(½abSinC) - Trigonometric graphs
- Pythagoras’ Theorem
- Hypotenuse

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