Question

In triangle $ABC$, $CD$ is the bisector of the angle $C$.

If $\cos \frac{C}{2}=\frac{1}{3}$ and $CD=6$ then find the value of $(\frac{1}{a}+\frac{1}{b})$

• A. $\frac{1}{9}$
• B. $\frac{4}{9}$
• C. $\frac{1}{3}$
• D. $\frac{2}{9}$

Answer 1

The correct option is A $\frac{1}{9}$

$Areaof△ABC=Areaof△ADC+Areaof△BDC\frac{1}{2}ab\sin c=\frac{1}{2}a\times CD\times \sin \frac{c}{2}+\frac{1}{2}b\times CD\times \sin \frac{c}{2}$

$ab\sin c=CD\sin \frac{c}{2}(a+b)\frac{a+b}{a-b}=\frac{\sin c}{CD\times \sin \frac{c}{2}}\Rightarrow \frac{2\sin \frac{c}{2}\cos \frac{c}{2}}{6\times \sin \frac{c}{2}}\Rightarrow \frac{2\times 1}{6\times 3}=\frac{1}{9}(\frac{1}{a}+\frac{1}{b})=\frac{1}{9}$

Hence $\frac{1}{9}$ is the correct answer.