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How to find angles of a triangle

Andrei

Member
Jan 18, 2013
36
We know that in triangle ABC angle A equals \(\displaystyle \alpha\) and side \(\displaystyle a=\frac{b+c}{2}.\) How to find angles B and C knowing that \(\displaystyle B\geqslant C\)? For which values of \(\displaystyle \alpha\) the problem has solutions?

ps. a, b, c are only notations.

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

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: how to find angles of a triangle

It seems you have given $\alpha$ in terms of $\alpha$.
 

Andrei

Member
Jan 18, 2013
36
Re: how to find angles of a triangle

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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: how to find angles of a triangle

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.
I think you are nearly there. Look at this picture, in which $AD$ is the angle bisector at $A$, and $BN$ is perpendicular to $AD$:



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 $AD$.​