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Can we apply the sines and cosines law on the external angles of triangle ?

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1 Answer 1

This answer assumes a triangle with angles $A, B, C$ with sides $a,b,c$.

Law of sines states that $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$$ Knowing the external angle is $\pi - \gamma$ if the angle is $\gamma$, $\sin(\pi-\gamma) = \sin \gamma$ because in the unit circle, you are merely reflecting the point on the circle over the y-axis and the sine value represents the $y$ value of the point, so the $y$ value will remain the same.
(Also, $\sin(\pi-\gamma) = \sin\pi \cos \gamma - \cos \pi \sin \gamma = \sin \gamma$), so $$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{\sin (\pi -A)}{a} = \frac{\sin (\pi -B)}{b} = \frac{\sin(\pi - C)}{c}$$

The law of cosines state that $$c^2=a^2+b^2-2ab\cos\gamma \Longrightarrow \cos \gamma = \frac{a^2 + b^2 - c^2}{2ab}$$ But $\cos(\pi-\gamma) = -\cos\gamma$ for pretty much the same reasons I used for the sine above. So if you are using external angles, the law of cosines will not work. However, if $\pi - \gamma$ is the external angle, then $$c^2=a^2+b^2+2ab\cos(\pi-\gamma) \Longrightarrow \cos(\pi- \gamma) = -\frac{a^2 + b^2 - c^2}{2ab}$$.

So in short, the answer is yes for law of sines, and no for law of cosines (unless you make the slight modification I made).

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