The basic thing about the sine formula is that the sides are proportional to the sines of the angles opposite to them.

We shall consider two different methods to derive the sine formula

### 1. Sine Formula from Acute-angled triangle

From the âˆ†ABC, draw a perpendicular AD to BC

In **âˆ† ABD:**

Sin B = \( \frac{AD}{c} \)

AD = cSinB……………(i)

In **âˆ† ACD: **

Sin C = \( \frac{AD}{b} \)

AD = bSinC…………….(ii)

**Equating (i) and (ii)**

cSinB = bSinC

âˆ´ \( \frac{SinB}{SinC} = \frac{b}{c} \rightarrow \frac{SinB}{b} = \frac{SinC}{c} \)

Similarly \( \frac{SinA}{SinB} = \frac{a}{b} \rightarrow \frac{SinA}{a} = \frac{SinB}{b} \)

and \( \frac{SinA}{SinC} = \frac{a}{C} \rightarrow \frac{SinA}{a} = \frac{Sinc}{c} \)

The three deductions are therefore combined as:

\( \frac{SinA}{a} = \frac{SinB}{b} = \frac{SinC}{c} \)

or \( \frac{a}{SinA} = \frac{b}{SinB} = \frac {c}{SinC} \)

### 2. Sine formula from obtuse-angled triangle

From the **âˆ†ABC**, draw a perpendicular AD to BC

In **âˆ†ABD:** Angles *âˆ ACB* and *âˆ ACD* are supplementary

**âˆ´ **AD = cSinB………….(i)

In **âˆ†ACD: **

AD = bSinC ………..(ii)

From the above

Sin ACD = Sin ABD = SinC

From i and ii

\( \frac{b}{c} = \frac{SinB}{SinC} \)Similarly \( \frac{a}{b} = \frac{SinA}{SinB} \)

and \( \frac{b}{c} = \frac{SinB}{SinC} \)

Combining the result then, \( \frac{SinA}{a} = \frac{SinB}{b} = \frac{SinC}{c} \)

Sine formula is used to solve problems connected with any triangle where most especially:

i) Two angles and one side are known (in other words, all the three angles and one side are known)

ii) Two sides and the angle opposite one of them (or non-included angle) are given.

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