Proving a/sinα=b/sinβ from Triangle Sides and Angles

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SUMMARY

The proof of the relationship a/sinα = b/sinβ is established using the height "h" of a triangle with sides "a", "b", and "c", and angles α and β opposite these sides. By applying the sine function, the equations b*sinα = h and a*sinβ = h are derived. This leads to the conclusion that b*sinα = a*sinβ, confirming the equality a/sinα = b/sinβ. The discussion confirms the validity of the proof presented by the user Kamataata.

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Kamataat
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I have to prove that [itex]a/\sin\alpha =b/\sin\beta[/itex]. A triangle has sides "a", "b", "c" and angles [itex]\alpha[/itex] and [itex]\beta[/itex] (opposite of the sides "a" and "b" respectively).

This is what I did:

Draw a line "h" as the height of the triangle on the side "c".

[tex]\sin\alpha = h/b[/tex]. Multiplying by "b" gives [tex]b\sin\alpha = h[/tex]

[tex]\sin\beta = h/a[/tex]. Multiplying by "a" gives [tex]a\sin\beta = h[/tex]

From this we see that [tex]b\sin\alpha = a\sin\beta[/tex] and from this
[tex]\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}[/tex].

Correct?

- Kamataata
 
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Thanks for the quick help!
 

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