Question

In a triangle $ABC$, with usual notation, $CD$ is the bisector of the angle $C.$ If $\cos \frac{C}{2}$ has the value $\frac{1}{2}$ and the length of $CD$ is $6$, 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. $\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{6}$


Area of $\Delta ABC=$ Area of $\Delta ACD+$ Area of $\Delta BCD$
$\Rightarrow \frac{1}{2}ab\sin C=\frac{1}{2}6b\sin \frac{C}{2}+\frac{1}{2}6a\sin \frac{C}{2}\Rightarrow ab\sin \frac{C}{2}\cos \frac{C}{2}=\frac{1}{2}6b\sin \frac{C}{2}+\frac{1}{2}6a\sin \frac{C}{2}\Rightarrow \frac{1}{2}ab\sin \frac{C}{2}=\frac{1}{2}6b\sin \frac{C}{2}+\frac{1}{2}6a\sin \frac{C}{2}\Rightarrow ab=6b+6a\Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{1}{6}$