<s>[INST] Prove that a triangle is right angled ... ... if one side is double the other and the angle opposite these sides differ by $\frac \pi 3$
It would be better if this was proven using only algebraic trigonometric identities(sine/cosine rules, etc.)
Let A,B,C be the angles and a,b,c be the respective sides opposite to them.
\begin{align} b &=2a \\
|B-A|&=\frac \pi 3 \\
\sin(B-A) &= \frac {\sqrt{3}} 2\\
\sin B \cos A-\cos B \sin A &=\frac {\sqrt{3}} 2\\
\frac b k \frac {b^2+c^2-a^2}{2bc}-\frac a k \frac {a^2+c^2-b^2}{2ac}&=\frac{\sqrt 3} 2 \\
\frac 1 {kc} (b^2-a^2)&=\frac {\sqrt 3} 2 \\
\frac {3a^2} {kc} &= \frac {\sqrt 3} 2
\end{align}
And I'm stuck at that.
Another approach
\begin{align}
\frac a {\sin A } &= \frac b {\sin B} \\
\frac {a}{\sin A} &= \frac {2a} {\sin (\frac \pi 3+A)}\\
\frac {\sqrt 3} 2\cos A-\frac 1 2 \sin A&= 2\sin A\\
\tan A&= \frac {\sqrt 3} 5\\
\end{align}
That doesn't lead anywhere too
 [/INST] Visual aid for in case you're interested in developing a synthetic proof:

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