Lesson Progress
0% Complete

The cosine formula is primarily used to solve triangles with the following dimensions:

(i) Two sides, given, with the angle between them given.

(ii) Three sides of the triangle are given.

Derivation of the Cosine Formula

Two cases of the derivation shall be considered as follows:

(i) Acute-angled triangle 

(ii) Obtuse-angled triangle

Screenshot 2022 06 08 at 03.56.56

Let BD = x

From Fig. i

CD = a – x

From Fig. ii

CD = x – a

In ∆ABD

AD2 = AB2 – BD2 (Pythagoras theorem)

AD2 = c2 – x2………….(i)

In ∆ACD:

AD2 = AC2 – CD2 (Pythagoras theorem)

from Fig. i
AD2 = b2 – (a – x)2……………(ii)

from Fig. ii
AD2 = b2 – (x – a)2

But (a – x)2 = (x – a)2

By equating i and ii

b2 – (a – x)2 = c2 – x2

b2 – a2 + 2ax – x2 = c2 – x2

Then 2ax = a2 + c2 – b2……………(iii)

But from ∆ABD 

CosB = \( \frac {x}{c} \)

x = cCosB

Substituting for x in equation iii

2acCos B = a2 + c2 – b2

Therefore,

\(\scriptsize Cos B = \normalsize \frac{a^2 \: + \: c^2 \: – \: b^2}{2ac}\)

Similarly, from ∆ACD 

Cos C = \( \frac{a \: -\:x}{b} \)

\( \scriptsize a \: -\:x = bCos\:C \)

\( \scriptsize -x = bCos\:C \: – \: a \)

\( \scriptsize x = a \: – \: bCos\:C \)

Substituting for x in equation iii

Then 2a(a – bCos C) = a2 + c2 – b2

2a2 – 2a(bCos C) = a2 + c2 – b2

-2abCos C = -2a2 + a2 + c2 – b2

-2abCos C = – a2 + c2 – b2

\( \scriptsize Cos \:C = \normalsize \frac{-a^2 \:+ \:c^2 \: – \: b^2}{-2ab} \)

Therefore,

\( \scriptsize Cos\:C = \normalsize \frac{a^2 \:+ \:b^2 \: – \: c^2}{2ab} \)

Simlarly,
\( \scriptsize Cos\:A = \normalsize \frac{b^2 \:+ \:c^2 \: – \: a^2}{2bc} \)

The above formula are suitable for calculating the various angles. The formulae may then be transformed through change of subject of formulae to derive the process of calculating the various sides as follows;

a2 = b2 + c2 – 2bcCosA

b2= a2 + c2 – 2acCosB

c2 = a2 + b2 – 2abCosC

Summary:

cosine rule

Responses

Your email address will not be published. Required fields are marked *

error: