Question

In $\Delta ABC,$ the median AD divides $\angle BAC$ such that $\angle BAD:\angle CAD=2:1.$ then $cos\frac{A}{3}$ is equal to

• A. $\frac{sinB}{2sinC}$
• B. $\frac{sinC}{2sinB}$
• C. $\frac{2sinB}{sinC}$
• D. None of these

Answer 1

The correct option is A

$\frac{sinB}{2sinC}$

$\frac{A}{3}=\angle CAD=\theta Now,(1+1)cot\alpha =1.cos2\theta -1.cot\theta \Rightarrow 2cot(B+2\theta )=cot2\theta -cot\theta \Rightarrow cot(B+2\theta )+cot\theta =cot2\theta -cot(B+2\theta )\frac{sin(B+3\theta )}{sin(B+2\theta ).sin\theta }=\frac{sinB}{sin(B+2\theta ).sin2\theta }\Rightarrow \frac{sin(B+A)}{sin\theta }=\frac{sinB}{sin2\theta }\Rightarrow sinC=\frac{sinB}{2cos\theta }\Rightarrow cos\frac{A}{3}=\frac{sinB}{2sinC}$