Given $\triangle ABC$ with $\alpha=2\beta$ and $b, c, a$ is forming an arithmetic sequence ($b$ is the $1$'st term, and $a$ is the last term), find $\alpha, \beta, \gamma$.
My attempts so far:
Let $c=b+k$ and $a=b+2k$. Then, from $\alpha=2\beta$, I got $\gamma = 180^\circ – 3\beta$.
With the Law of Sines , I can write
$$\dfrac{b}{\sin \beta}=\dfrac{b+k}{\sin 3\beta}=\frac{b+2k}{\text{sin } 2\beta}$$
From $\dfrac{b}{\sin\beta}=\dfrac{b+k}{\sin 3\beta}$, it gives $\cos 2\beta = \dfrac{k}{2b}$.
From $\dfrac{b}{\sin \beta}=\dfrac{b+2k}{\sin 2\beta}$, it gives $\cos \beta = \dfrac{b+2k}{2b}$.
After this, I don't know what should I do. Is there any other theorem that can be used to solve this problem?
Best Answer
From $b,c,a$ forming an arithmetic sequence, $$c-b=a-c$$ $$2c=a+b$$ Applying sine rule, $$2\sin C=\sin A+\sin B$$ Using the given data $A=2B$, we can rewrite the above as, $$2\sin 3B=\sin2B+\sin B$$ $$\require{cancel}2\cdot \cancel{2\sin\frac{3B}{2}}\cos\frac{3B}2=\cancel{2\sin\frac{3B}{2}}\cos\frac B2\qquad\small{(\because\sin\frac{3B}{2}\neq0})$$ Now letting $\cos\frac B2=x$, we get, $$2(4x^3-3x)=x$$ $$x^2=\frac78$$ Therefore, $$\cos B=2\cos^2\frac B2-1=\frac34$$ $$B=\cos^{-1}\frac34$$
So we have found $B$ alias $\beta$ which leads to find other angles as well.