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Sine and cosine rules - Higher

Sine and cosine rules - Higher

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Summary

Sine and cosine rules

In a nutshell

The sine and cosine rules can be used to work out a missing side or angle in any triangle. Similarly, the trigonometric area of a triangle can be used to find the area of a triangle without being given the perpendicular height.


Labelling a triangle

It is common convention to label the vertices (and corresponding angle) of a triangle with capital letters.  The side opposite the angle is labelled with its lowercase counterpart. Here is a diagram to illustrate:


Maths; Trigonometry; KS4 Year 10; Sine and cosine rules - Higher



The sine rule

When is it used?

The sine rule is used when you have an angle-side pair (e.g. side BB​ and angle bb​), and either another side or another angle.


The sine rule formula

There are two forms of the sine rule: the side form and the angle form.


Side form

Angle form

asin(A)=bsin(B)=csin(C)\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}​​
sin(A)A=sin(B)b=sin(C)c\frac{\sin(A)}{A}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}​​


Example 1

In the diagram above, the angles AA​ and BB​ are 4040^\circ and 7070^\circ respectively. If the side aa​ is 5cm5cm​ long, what is the length of side bb​ to the nearest integer?


You are given AA​, BB​, aa and you are told to find the value of bb. So, use the first 2 fractions of the sine rule. Use the side form as you need to find a length:

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


Substitute the known values:

5sin(40)=bsin(70)\frac{5}{\sin(40)}=\frac{b}{\sin(70)}


Solve for bb​:

b=sin(70)×5sin(40)=7.309511...b = \sin(70)\times \frac{5}{\sin(40)} = 7.309511...​​


b=7cm\underline{b =7cm} to the nearest integer.



The cosine rule

When is it used?

The cosine rule is used when you have three sides or two sides and the angle between them.


The cosine rule formula

Again, there are two forms of the cosine rule: the side form and the angle form.


​​Side form

Angle form

a2=b2+c22bccos(A)a^2=b^2+c^2-2bc\cos(A)​​
cos(A)=b2+c2a22bc\cos(A)=\frac{b^2+c^2-a^2}{2bc}​​


Tip: If you're confident enough in your algebra and rearranging, you can just memorise the side form.


Example 2

The lengths of the sides of a triangle are 5cm5cm, 7cm7cm​, and 10cm10cm​. What is the size of the angle opposite the side with length 7cm7cm​ to 1 decimal place?


Always start off by drawing and labelling a sketch if you're not given one:


Maths; Trigonometry; KS4 Year 10; Sine and cosine rules - Higher


Note: Labelling correctly is very important when using the cosine rule. The angle form of the cosine rule has the angle AA​, so make sure the side with length 7cm7cm​ is labelled as aa​. It doesn't matter which of the remaining sides you label bb​ and cc​.


Now, use the angle form of the cosine rule:

cos(A)=b2+c2a22bc\cos(A)=\frac{b^2+c^2-a^2}{2bc}​​


Substitute in known values:

cos(A)=102+52722(5)(10)\cos(A)=\frac{10^2+5^2-7^2}{2(5)(10)}​​


cos(A)=0.76\cos(A)=0.76​​


Use the inverse cosine function to get the angle by itself:

A=cos1(0.76)=40.535802...A=\cos^{-1}(0.76)=40.535802...​​


A=40.5° (1 d.p.)\underline{A=40.5\degree \ (1 \ d.p.)}



Area of a triangle

When is it used?

This is used when you have two sides and the angle between them, and want to find the area of the triangle.


The formula for the area of a triangle

Area=12absin(C)Area=\frac{1}{2}ab\sin(C)​​


Example 3

A triangle has sides 2cm2cm​, 4cm4cm​ and the angle between them is 2424^\circ. Find the area of the triangle to 2 decimal places.


Sketch the triangle first and label the sides and angles.


Maths; Trigonometry; KS4 Year 10; Sine and cosine rules - Higher


Note: Again, it's important to label your triangle properly. The formula has sin(C)\sin(C), so you know that your given angle needs to be CC​, which means the side opposite should be labelled as cc​.


Now, use the formula:

Area=12absin(C)Area=\frac{1}{2}ab\sin(C)​​


Substitute in known values:

Area=12(2)(4)sin(24)Area = \frac{1}{2}(2)(4)\sin(24)​​


Area=4×sin(24)=1.62694....Area =4\times \sin(24)=1.62694....​​


Area =1.63cm2 (2 d.p.)\underline{Area\,=1.63cm^2 \ (2 \ d.p.)}


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FAQs - Frequently Asked Questions

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