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ps. a, b, c are only notations.

answer. \(\displaystyle \frac{\pi-\alpha}{2}\pm\arccos(2\sin\frac{\alpha}{2})\)

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ps. a, b, c are only notations.

answer. \(\displaystyle \frac{\pi-\alpha}{2}\pm\arccos(2\sin\frac{\alpha}{2})\)

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Here is what I do. I express a through b, c and \(\displaystyle \alpha\) by cosine theorem. So I have an equation with only b and c as unknown. I obtain \(\displaystyle \frac{b}{c}\) by sine theorem and I substitute this ratio in the first equation. I also know that \(\displaystyle \alpha=\pi-B-C.\) So I have two equations with two unknowns. But it is hard to solve.

I also noticed that the bisector of angle A divide side a in two segments: b/2 and c/2.

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- Feb 7, 2012

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I think you are nearly there. Look at this picture, in which $AD$ is the angle bisector at $A$, and $BN$ is perpendicular to $AD$:Here is what I do. I express a through b, c and \(\displaystyle \alpha\) by cosine theorem. So I have an equation with only b and c as unknown. I obtain \(\displaystyle \frac{b}{c}\) by sine theorem and I substitute this ratio in the first equation. I also know that \(\displaystyle \alpha=\pi-B-C.\) So I have two equations with two unknowns. But it is hard to solve.

I also noticed that the bisector of angle A divide side a in two segments: b/2 and c/2.

The angle at $B$ is obviously $\angle ABN + \angle NBD$. Equally obviously, $\angle ABN = \frac{\pi}2 - \frac\alpha2$, so we just need to show that $\cos(\angle NBD) = 2\sin\bigl(\frac\alpha2\bigr)$. But $\cos(\angle NBD) = \frac{BN}{BD}$, and you have already shown that $BD = c/2$, so you just need to observe that $BN = c\sin\bigl(\frac\alpha2\bigr)$, which is evident from the triangle $ABN$.

You can get the result for the angle at $C$ in a similar way by dropping a perpendicular from $C$ to the extension of $AD$.

You can get the result for the angle at $C$ in a similar way by dropping a perpendicular from $C$ to the extension of $AD$.