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The sine rule

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 bbb​ and angle BBB​), and either another side or another angle. 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)\dfrac{a}{\sin(A)}=\dfrac{b}{\sin(B)}=\dfrac{c}{\sin(C)}sin(A)a​=sin(B)b​=sin(C)c​​​

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


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


Maths; Trigonometry; KS5 Year 12; The sine rule


Example 1

A triangle has angles AAA​ and BBB which are 40∘40^\circ40∘ and 70∘70^\circ70∘ respectively. If the side aaa​ is 5cm5cm5cm​ long, what is the length of side bbb​ to 111 decimal place?


Use the sine rule to form an equation involving the given values of AAA​, BBB​, aaa and the unknown value of bbb: 

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


Substitute the known values:

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


Solve for bbb​:

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

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


Example 2

In △ABC\triangle ABC△ABC, AB=12 cmAB = 12 \ cmAB=12 cm, BC=15 cmBC=15 \ cmBC=15 cm and ∠CAB=75∘\angle CAB = 75 ^{\circ}∠CAB=75∘. Find angle BBB and CCC.​

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}asin(A)​=bsin(B)​=csin(C)​​​


Substitute the known values:

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


Solve for angle BBB:

​12sin⁡(75)15=sin⁡(B)\dfrac{12\sin(75)}{15}= \sin(B)1512sin(75)​=sin(B)​​


B=sin⁡−1(12sin⁡(75)15)=50.60063...B= \sin^{-1}(\dfrac{12\sin(75)}{15}) =50.60063...B=sin−1(1512sin(75)​)=50.60063...

​​

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


Angles in a triangle add up to 180∘180^{\circ}180∘. Use this to find angle CCC:

180−50.6−75= 180 -50.6 -75=180−50.6−75=​​

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



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Finding missing lengths in a triangle

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The sine rule

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FAQs - Frequently Asked Questions

What is the sine rule used for?

The sine rule is used to work out the missing lengths or angles in any type of triangle.

What is the side form of the sine rule?

The side form of the sine rule is: a/sin(A) = b/sin(B) = c/sin(C).

What is the angle form of the sine rule?

The angle form of the sine rule is: sin(A)/a = sin(B)/b = sin(C)/c.

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