Question

If in a $\Delta ABC,\frac{cosA+2cosC}{cosA+2cosB}=\frac{sinB}{sinC}$, then the triangle is

• A. Isosceles
• B. Right angled
• C. Isosceles right angled
• D. None of these

Answer 1

Answer: (C)

Isosceles right angled

$\frac{\cos A+2\cos C}{\cos A+2\cos B}=\frac{\sin B}{\sin C}$

$\Rightarrow \sin C(\cos A+2\cos C)=\sin B(\cos A+2\cos B)$

$\Rightarrow \cos A\sin C+2\sin C\cos C=\sin B\cos A+2\sin B\cos B$

$\Rightarrow \cos A(\sin C-\sin B)+\sin 2C-\sin 2B=0$

$\Rightarrow \cos A2\cos \left(\frac{B+C}{2}\right)\sin \left(\frac{B-C}{2}\right)+2\cos (B+C)\sin (B-C)=0$

$\Rightarrow \cos A\cos (\frac{\pi }{2}-\frac{A}{2})\sin \left(\frac{B-C}{2}\right)+\cos (B+C)\sin (B-C)=0$

$\Rightarrow \cos A\sin \frac{A}{2}\sin \left(\frac{B-C}{2}\right)+\cos (\pi -A)2\sin \left(\frac{B-C}{2}\right)\cos \left(\frac{B-C}{2}\right)=0$

$\Rightarrow \cos A\sin \frac{A}{2}\sin \left(\frac{B-C}{2}\right)-\cos A\times 2\sin \left(\frac{B-C}{2}\right)\cos \left(\frac{B-C}{2}\right)=0$

$\Rightarrow \cos A\sin \left(\frac{B-C}{2}\right)[\sin \frac{A}{2}-2\cos \left(\frac{B-C}{2}\right)]=0$

So, either $\cos A=0=\cos \frac{\pi }{2}$ or $\sin \left(\frac{B-C}{2}\right)=0$

$\Rightarrow A=\frac{\pi }{2}$ or $B=C$

Hence the triangle is isosceles right angled triangled.