Sides of a triangle are in Arithmetic Progression, then find $\tan (\alpha+ \frac{\beta}{2})$

Question

The sides of a triangle are in Arithmetic Progression. If the smallest angle of the triangle is $\alpha$ and largest angle of the triangle exceeds the smallest angle by $\beta$, then find the value of $\tan (\alpha+ \frac{\beta}{2})$

Would it be correct to assume sides of triangle of as $1,2,3$ and then apply cosine rule to find angles? Or could someone propose a better approach?

Answer 1

Here are some hints.

I don't think you can assume the sides of the triangle are $1,2,3$. You can assume that they're $x, x+c,$ and $x+2c$, though.

If the smallest and largest angles are $\alpha$ and $\alpha + \beta$, then you know the middle angle is $\pi - 2 \alpha - \beta$.

Then from Law of Sines you know that

$$\frac{x}{\sin \alpha} = \frac{x+c}{\sin (2\alpha + \beta)} = \frac{x+2c}{\sin (\alpha+\beta)}.$$

(I used the fact that $\sin(\pi - \theta) = \sin(\theta)$.)

Now, $(2 \alpha + \beta) = 2(\alpha + \frac{\beta}{2})$, so using a half-angle formula along the way could get you somewhere.

Can you take it from here?