Question

If $A(0,0),B(\cos \alpha ,\sin \alpha )$ and $C(\cos \beta ,\sin \beta )$ represent the vertices of a right angled triangle, then which of the following is/are correct?

• A. $\sin \left(\frac{\alpha -\beta }{2}\right)=\frac{1}{\sqrt{2}}$
• B. $\cos \left(\frac{\alpha -\beta }{2}\right)=-\frac{1}{\sqrt{2}}$
• C. Coordinates of orthocentre are $(0,0)$
• D. Coordinates of orthocentre are $(\cos \alpha ,\sin \alpha )$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\sin \left(\frac{\alpha -\beta }{2}\right)=\frac{1}{\sqrt{2}}$
B $\cos \left(\frac{\alpha -\beta }{2}\right)=-\frac{1}{\sqrt{2}}$
C Coordinates of orthocentre are $(0,0)$

Given : $A(0,0),B(\cos \alpha ,\sin \alpha )$ and $C(\cos \beta ,\sin \beta )$
Finding the lengths of the triangle,
$AB=\sqrt{{\cos }^2\alpha +{\sin }^2\alpha }=1\text{units}$
$AC=\sqrt{{\cos }^2\beta +{\sin }^2\beta }=1\text{units}$
$BC=\sqrt{(\cos \alpha -\cos \beta {)}^2+(\sin \alpha -\sin \beta {)}^2}\Rightarrow BC=\sqrt{2-2(\cos \alpha \cos \beta +\sin \alpha \sin \beta )}\Rightarrow BC=\sqrt{2-2\cos \left(\alpha -\beta \right)}\text{units}$

As $AC=AB$, triangle is right angled isosceles triangle. So, hypotenuse is $BC$
Now, using Pythagoras theorem, we get
$A{B}^2+A{C}^2=B{C}^2$
$\Rightarrow 2=2-2\cos \left(\alpha -\beta \right)$
$\Rightarrow \cos \left(\alpha -\beta \right)=0\Rightarrow \alpha -\beta =\frac{(2n+1)\pi }{2},n\in Z\therefore \sin \left(\frac{\alpha -\beta }{2}\right)=\cos \left(\frac{\alpha -\beta }{2}\right)=\pm \frac{1}{\sqrt{2}}$

$△ABC$ is right angled at $A$, so the coordinates of orthocentre are same as coordinates of $A$, i.e., $(0,0)$