# TRIGONOMETRY: SINE AND COSINE RULE

CONTENT

- Derivation and application of sine rule.
- Derivation and application of cosine rule.

**SINE RULE**

Given any triangle ABC (acute or obtuse), with the angles labelled with capital letters A, B, C and the sides opposite these angles labelled with the corresponding small letters a, b, and c respectively as shown below.

The sine rule states that;

\(\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}\)

OR

\(\frac{sinA}{a} = \frac{sinB}{b} = \frac{sinC}{c}\)

**PROOF OF THE RULE**

**Using Acute – angled triangle**

** Given:** Any ∆ABC with B acute.

** To prove: **\(\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}\)

** Construction:** Draw the perpendicular from C to AB.

*Proof: **Using the lettering in the diagram above.*

\(sinA = \frac{h}{b} \\ h = bsinA …….(i) \\ sinB = \frac{h}{a} \\ h = asinB …….(ii)\)

From equation (i) and (ii)

\(bsinA = asinB \\ ∴ \frac{a}{sinA} = \frac{b}{sinB}\)

Similarly, by drawing a perpendicular from B to AC

\(\frac{a}{sinA} = \frac{c}{sinC} \\ ∴ \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC} \)

**Q.E.D**

*Using Obtuse – angled triangle*

** Given:** any ∆ABC with B obtuse

** To Prove: **\(\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}\)

** Construction:** Draw the perpendicular from C to AB produced.

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