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The sine rule

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Tutor: Labib

Summary

The sine rule

In a nutshell

The sine rule is used to work out the missing lengths or angles in any type of triangle. There are two forms of the sine rule you will need to know.



The sine rule formula

The sine rule is used when you have a side-angle pair (e.g. side bb​ and angle BB​), and either another side or another angle. There are two forms of the sine rule: The side form and the angle form.

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Side form

Angle form

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

sin(A)a=sin(B)b=sin(C)c\dfrac{\sin(A)}{a}=\dfrac{\sin(B)}{b}=\dfrac{\sin(C)}{c}​​


This is for a triangle with sides a,b,ca,b,c​ and corresponding angles A,B,CA,B,C.​


Maths; Trigonometry; KS5 Year 12; The sine rule


Example 1

A triangle has angles AA​ and BB which are 4040^\circ and 7070^\circ respectively. If the side aa​ is 5cm5cm​ long, what is the length of side bb​ to 11 decimal place?


Use the sine rule to form an equation involving the given values of AA​, BB​, aa and the unknown value of bb

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


Substitute the known values:

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


Solve for bb​:

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

b=7.3 cm (1d.p.)\underline{b =7.3 \ cm \ (1d.p.)}


Example 2

In ABC\triangle ABCAB=12 cmAB = 12 \ cm, BC=15 cmBC=15 \ cm and CAB=75\angle CAB = 75 ^{\circ}. Find angle BB and CC.

Maths; Trigonometry; KS5 Year 12; The sine rule


To solve angles use the angle form of the sine rule:

sin(A)a=sin(B)b=sin(C)c\dfrac{\sin(A)}{a}=\dfrac{\sin(B)}{b}=\dfrac{\sin(C)}{c}​​


Substitute the known values:

sin(75)15=sin(B)12\dfrac{\sin(75)}{15} = \dfrac{\sin(B)}{12}​​


Solve for angle BB:

12sin(75)15=sin(B)\dfrac{12\sin(75)}{15}= \sin(B)​​


B=sin1(12sin(75)15)=50.60063...B= \sin^{-1}(\dfrac{12\sin(75)}{15}) =50.60063...

​​

B=50.6(1d.p.)\underline{B=50.6^{\circ}(1d.p.)}​​


Angles in a triangle add up to 180180^{\circ}. Use this to find angle CC:

18050.675= 180 -50.6 -75=​​

C=54.4(1.dp) \underline{C=54.4^{ \circ} (1.dp)}​​



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FAQs - Frequently Asked Questions

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