Cambridge International Mathematics

Proof(for a triangle with acute angleA):

Consider triangle ABC shown.

Using Pythagoras’ theorem, we find

b^2 =h^2 +x^2 ,soh^2 =b^2 ¡x^2

and a^2 =h^2 +(c¡x)^2

Thus, a^2 =(b^2 ¡x^2 )+(c¡x)^2

) a^2 =b^2 ¡x^2 +c^2 ¡ 2 cx+x^2

) a^2 =b^2 +c^2 ¡ 2 cx ::::::(1)

But in¢ACN, cosA=

x

b

and so x=bcosA

So, in (1), a^2 =b^2 +c^2 ¡ 2 bccosA

Similarly, we can show the other two equations to be true.

Challenge: Prove theCosine Rule a^2 =b^2 +c^2 ¡ 2 bccosA, in the case whereAis an obtuse angle.

You will need to use cos(180o¡μ)=¡cosμ:

We use thecosine rulewhen we are given:

² two sidesand theincluded anglebetween them, or

² three sides.

Useful rearrangements of the cosine rule are:

cosA=

b^2 +c^2 ¡a^2

2 bc

, cosB=

a^2 +c^2 ¡b^2

2 ac

, cosC=

a^2 +b^2 ¡c^2

2 ab

They can be used if we are given all three side lengths of a triangle.

Example 8 Self Tutor

Find, correct to 2 decimal places,

the length of BC.

By the cosine rule:

a^2 =b^2 +c^2 ¡ 2 bccosA

) a=

p

122 +10^2 ¡ 2 £ 12 £ 10 £cos 38o

) a¼ 7 : 41

) BC is 7 : 41 m in length.

AB

C

A

ba

xcx¡-¡

h

N

A C

B

10 m

38°

12 m

A C

B

10 m

38°

12 m

am

Further trigonometry (Chapter 29) 589

GEOMETRY

PACKAGE

IGCSE01

cyan magenta yellow black

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(^05255075950525507595)
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Y:\HAESE\IGCSE01\IG01_29\589IGCSE01_29.CDR Friday, 31 October 2008 9:54:56 AM TROY