Question

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

• A. equilateral
• B. isosceles
• C. right angled
• D. obtuse angled

Answer 1

Answer: (B), (C)

The correct options are
B isosceles
C right angled

The given relation can be written as

$sinCcosA+2cosCsinC=sinBcosA+2cosBsinB$

$cosA(sinB-sinC)+sin2B-sin2C=0$

or $cosA(sinB-sinC)+2cos(B+C)sin(B-C)=0$

or $cosA(sinB-sinC)+2cos({180}^o-A)sin(B-C)=0$

or $cosA[sinB-sinC-2sin(B-C\left)\right]=0$

from which it follows that either $cosA=0$, so that $A=\pi /2$ and $\Delta$ABC is right-angled, or

$sinB-sinC-2sin(B-C)=0$

$\Rightarrow (b-c)-2(bcosC-ccosB)=0$
[by the law of sines]

$\Rightarrow (b-c)-2(\frac{a^2+b^2-c^2}{2a}-\frac{c^2+a^2-b^2}{2a})=0$ [cosine rule]

$\Rightarrow a(b-c)-2(b^2-c^2)=0$

$\Rightarrow (b-c)[a-2(b+c\left)\right]=0$

$\Rightarrow b-c=0(\because b+c>a)$

$\Rightarrow$ triangle is isosceles