Proof:

 

Begin by constructing altitude h1 from vertex B to side b. Now note that

,

so

h1 = csinA.

Similarly,

and

h1 = asinC.

Thus,

csinA = asinC

and

.

Obviously, similar reasoning can be used to show that

and

.

Therefore, we have demonstrated that

Considering the diagram we used in the proof, we have actually only verified the law of sines for acute triangles. What happens if the triangle is obtuse?

Begin by extending side AC and constructing the altitude from vertex B.

Now notice that angles ACB and BCD are supplementary. (Let angle ACB = g and angle BCD = d.Thus,

sing = sind,

implying

sind = h1/a = sing.

Solving for h1,

h1 = asing.

Also, note that sinA = h1/c, so h1 = csinA.

Thus,

asing = csinA

and

a/(sinA) = c/(sing).

Similar reasoning can be used to obtain the rest of the derivation.

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