# The Sine Law

Let ABC be a triangle, acute angled in fig(i), obtuse angled in fig(ii) and right angled in fig(iii).

A perpendicular AD is drawn on BC [BC is produced to D in fig. (ii) and D coincides C in fig (iii)].

To Prove:

Proof:**1. When C is acute-angled as shown in fig(i),**From triangle ADB,

From triangle ADC,

From eq^{n}:(i) & eq^{n}:(ii)

2. When C is obtuse-angled as shown in fig(ii),

From triangle ADB,

From triangle ADC,

From eq^{n}:(i) & eq^{n}:(ii)

3. When C is right-angled as shown in fig(iii),

Also,

From eq^{n}:(i) & eq^{n}:(ii)

Hence, for all cases, we have,

..........(I)

Similarly, when a perpendicular is drawn from B on CA, we have

..........(II)

Combining eq^{n}:(I) & eq^{n}:(II)

For all cases, we have,

, which is the Sine Law.