How to find angles of a triangle

[Andrei1]

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]

Re: how to find angles of a triangle

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

 

[Andrei1]

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]

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 $AD$.​

 

[CellCoree]

To find the angles of a triangle, we can use the law of cosines or the law of sines. In this case, since we are given the value of angle A and the relationship between the sides, we can use the law of cosines.

According to the law of cosines, we know that in a triangle with sides a, b, and c and angle A opposite of side a, the following equation holds: a^2 = b^2 + c^2 - 2bc\cos A.

Substituting in the given values, we get:

\left(\frac{b+c}{2}\right)^2 = b^2 + c^2 - 2bc\cos \alpha

Simplifying, we get:

b^2 + c^2 + 2bc - 4bc\cos \alpha = 0

Using the quadratic formula, we can solve for b or c. However, since we are given that B\geqslant C, we can assume that c is the longer side and b is the shorter side. Therefore, we can solve for b using the negative root.

b = c(2\cos \alpha - 1)

Now, to find angles B and C, we can use the law of sines. According to the law of sines, in a triangle with sides a, b, and c and angles A, B, and C opposite of their respective sides, the following equation holds: \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}.

Substituting in the values we have found for a and b, we get:

\frac{\sin \alpha}{\frac{b+c}{2}} = \frac{\sin B}{b}

Simplifying, we get:

\sin B = \frac{2\sin \frac{\alpha}{2}}{2\cos \alpha - 1}

Using the inverse sine function, we can solve for angle B. Similarly, we can solve for angle C using the same process.

Therefore, the solution for angles B and C is:

B = \arcsin \left( \frac{2\sin \frac{\alpha}{2}}{2\cos \alpha - 1} \right)

C = \arcsin \left( \frac{2\sin \frac{\alpha}{2}}{