Question

In a triangle ABC, CD is the bisector of the angle C. If $\cos \frac{C}{2}$ has the value $\frac{1}{3}$ and length of $CD=6$, then$(\frac{1}{a}+\frac{1}{b})$ has the value equal to

• A. $\frac{1}{9}$
• B. $\frac{1}{12}$
• C. $\frac{1}{6}$
• D. none

Answer 1

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

In $\Delta ADC,$

$cos\frac{C}{2}=\frac{CD}{b}=\frac{6}{b}=\frac{1}{3}$ (given)

$\Rightarrow b=18$

In $\Delta BCD\Rightarrow cos\frac{C}{2}=\frac{CD}{a}\Rightarrow \frac{1}{3}=\frac{6}{a}$

$\Rightarrow a=18$

$\therefore \frac{1}{a}+\frac{1}{b}=\frac{1}{18}+\frac{1}{18}=\frac{2}{18}=\frac{1}{9}$