Question

If the three points $(a,b),(a+r\cos \alpha ,b+r\sin \alpha ),(a+r\cos \beta ,b+r\sin \beta )$ are the vertices of an equilateral triangle. Find the value of $|\alpha -\beta |$

Answer 1

Let $A(a,b);B(a+r\cos \alpha ,b+\sin \alpha );C(a+r\cos \beta ,b+r\sin \beta )$

We have

$\Rightarrow AC=\sqrt{(a+r\cos \beta -a{)}^2+(b+r\sin \beta -b{)}^2}$

$\Rightarrow AC=\sqrt{\left(r^2{\cos }^2\beta \right)+\left(r^2{\sin }^2\beta \right)}=\sqrt{r^2}=r$

$\Rightarrow AC=r$

Now,

$\Rightarrow BC=\sqrt{(a+r\cos \alpha -a-r\cos \beta {)}^2+(b+r\sin \alpha -b-r\sin \beta {)}^2}$

$\Rightarrow BC=\sqrt{r^2(\cos \alpha -\cos \beta {)}^2+r^2(\sin \alpha -\sin \beta {)}^2}$

$\Rightarrow BC=r\sqrt{({\cos }^2\alpha +{\cos }^2\beta -2\cos \alpha \cos \beta )+({\sin }^2\alpha +{\sin }^2\beta -2\sin \alpha \sin \beta )}$

$\Rightarrow BC=r\sqrt{1+1-2\cos \alpha \cos \beta -2\sin \alpha \sin \beta }$

$\Rightarrow BC=r\sqrt{2-2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )}$

$\Rightarrow BC=r\sqrt{2-2\cos (\alpha -\beta )}$

As $△ABC$ is equilateral triangle, $AC=BC$

$\Rightarrow r\sqrt{2-2\cos (\alpha -\beta )}=r$

$\Rightarrow \sqrt{2-2\cos (\alpha -\beta )}=1$

$\Rightarrow 2-2\cos (\alpha -\beta )=1$

$\Rightarrow \cos (\alpha -\beta )=\frac{1}{2}$

Therefore, $|\alpha -\beta |=\frac{\pi }{3}$