Question

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

• A. $3$
• B. $6$
• C. $9$
• D. $12$

Answer 1

The correct option is C $9$


Given: $CD=6,\cos \frac{C}{2}=\frac{1}{3}$
Let $\Delta$ be the area of triangle $ABC$. ${\Delta }_1\text{and}{\Delta }_2$ be the areas of smaller triangles. Then we have:
$\Delta ={\Delta }_1+{\Delta }_2$
$\Rightarrow \frac{1}{2}ab\sin C=\frac{1}{2}6b\sin \frac{C}{2}+\frac{1}{2}6a\sin \frac{C}{2}$
$\Rightarrow \frac{1}{2}ab\times 2\sin \frac{C}{2}\cos \frac{C}{2}=3b\sin \frac{C}{2}+3a\sin \frac{C}{2}$
$\Rightarrow ab\cos \frac{C}{2}=3b+3a$
$\Rightarrow ab\times \frac{1}{3}=3b+3a$
$\Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{1}{9}\Rightarrow \frac{1}{\lambda }=\frac{1}{9}\Rightarrow \lambda =9$