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$\begingroup$

Sine rule:

$$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{C}{\sin(C)}=2R$$

But I want to know what is

$$\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}=?$$

On wikipedia it says it is equal to $\dfrac{2\Delta}{abc}$ shouldn't it be simply $\frac 1{2R}$ ?

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    You've got it the wrong way: $$\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}=\frac{1}{2R}$$ – 2017-02-17
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    Mine [wikipedia](https://en.wikipedia.org/wiki/Law_of_sines) says different (no $\Delta$ indeed). And $\Delta\neq \frac{1}{\nabla}$. – 2017-02-17
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    ok and you can use $$\sin(\alpha)=\frac{a}{2R}$$ etc – 2017-02-17

2 Answers 2

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There is no contradiction: $$\dfrac{2\Delta}{abc} = \dfrac1{2R}$$ i.e. the area of the triangle is $$\Delta = \dfrac{abc}{4R}$$ where $R$ is the radius of the circumcircle.

For example, with a $3,4,5$ right angled triangle, the area is $6$ and the circumcircle radius is $\frac52$ and you have $$\dfrac1{2R}=\dfrac1{2 \times \frac52}=\dfrac15=\dfrac{2 \times 6}{3\times 4 \times 5} = \dfrac{2\Delta}{abc}$$

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$R = \frac{a}{2 \sin A} = \frac{abc}{2 bc \sin A} = \frac{abc} { 4 \Delta} $. Now $ \frac{1}{2R} = \frac{2 \Delta}{abc} $