Question

In a triangle, the line joining the circumcentre to the incentre is parallel to $BC$, then $\cos B+\cos C$ is equal to

• A. $\frac{3}{2}$
• B. $1$
• C. $\frac{3}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (B)

$1$

$M$ is the mid-point of $BC$
$\therefore MC=\frac{a}{2},OC=R,OM=ID=r$
$\therefore △OMC,R^2=r^2+\frac{a^2}{4}$
$=r^2+\frac{{\left(2R\sin A\right)}^2}{4}$
$=r^2+{\left(R\sin A\right)}^2$
$\Rightarrow R^2(1-{\sin }^{2}A)=r^2$
$\Rightarrow R^2{\cos }^{2}A=r^2$
$\Rightarrow {\cos }^{2}A=\frac{r^2}{R^2}$
$\Rightarrow r=R\cos A$
$\because \cos A+\cos B+\cos C$
$=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)+1-2{\sin }^2\frac{C}{2}$
$=1+2\cos (\frac{\pi }{2}-\frac{C}{2})\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}$
$=1+2\sin \left(\frac{C}{2}\right)\cos \left(\frac{A-B}{2}\right)-2{\sin }^2\frac{C}{2}$
$=1+\sin \left(\frac{C}{2}\right)(\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{C}{2}\right))$
$=1+\sin \left(\frac{C}{2}\right)(\cos \left(\frac{A-B}{2}\right)-\sin (\frac{\pi }{2}-\frac{A+B}{2}))$
$=1+\sin \left(\frac{C}{2}\right)(\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right))$
Using transformation angle formula, we get
$=1+\sin \left(\frac{C}{2}\right)2\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)$
$=1+4\sin \left(\frac{A}{2}\right)\sin \left(\frac{B}{2}\right)\sin \left(\frac{C}{2}\right)$
$=1+\frac{r}{R}=1+\cos A$
$\therefore \cos B+\cos C=1$