Question

The distance between the circum-centre and the orthocentre of a $△ABC$ is

• A. $R\sqrt{1-8\cos A\cos B\cos C}$
• B. $R\sqrt{1-8\sin A\cos B\cos C}$
• C. $R\sqrt{1-8\cos A\sin B\cos C}$
• D. $R\sqrt{1-8\sin A\sin B\sin C}$

Answer 1

Answer: (A)

$R\sqrt{1-8\cos A\cos B\cos C}$

Let O be the Cirumcentre and H be the orthocentre then

$OH=\sqrt{9R^2-(a^2+b^2+c^2)}$

$=\sqrt{9R^2-4R^2({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C)}$

$=R\sqrt{9-4({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C)}$

$=R\sqrt{9-4\left(\frac{1-\cos 2A+1-\cos 2B+1-\cos 2C}{2}\right)}$

$=R\sqrt{9-2(3-(\cos 2A+\cos 2B+\cos 2C\left)\right)}$

$=R\sqrt{3+2(\cos 2A+\cos 2B+\cos 2C))}$

$=R\sqrt{3+2\left(2\cos \right(A+B\left)\cos \right(A-B)+2{\cos }^{2}C-1)}$

$=R\sqrt{3+2\left(2\cos \right(\pi -C\left)\cos \right(A-B)+2{\cos }^{2}C-1)}$

$=R\sqrt{3+2(-2\cos (C\left)\cos \right(A-B)+2{\cos }^{2}C-1)}$

$=R\sqrt{3+2\left(2\cos \right(C\left)\right(\cos C-\cos (A-B))-1)}$

$=R\sqrt{3+4\cos \left(C\right)(\cos C-\cos (A-B\left)\right)-2}$

$=R\sqrt{1+4\left(\cos \right(C\left)\right(\cos (\pi -(A+B\left)\right)-\cos (A-B))}$

$=R\sqrt{1-4\left(\cos \right(C\left)\right(\cos (A+B)+\cos (A-B)\left)\right)}$

$=R\sqrt{1-4\left(\cos \right(C\left)\right(2\cos A\cos B\left)\right)}$

$=R\sqrt{1-8\cos C.\cos A.\cos B}$