In a triangle ABC,CD is the bisector of the a | vedclass

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(A) Let $CD = l = 6$. The area of $\Delta ABC$ is the sum of the areas of $\Delta ACD$ and $\Delta BCD$.$	ext{Area}(\Delta ABC) = 	ext{Area}(\Delta

Vedclass · Accepted Answer

$\frac{1}{9}$

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(A) Let $CD = l = 6$. The area of $\Delta ABC$ is the sum of the areas of $\Delta ACD$ and $\Delta BCD$.$	ext{Area}(\Delta ABC) = 	ext{Area}(\Delta ACD) + 	ext{Area}(\Delta BCD)$$\frac{1}{2} ab \sin C = \frac{1}{2} b l \sin \frac{C}{2} + \frac{1}{2} a l \sin \frac{C}{2}$Using $\sin C = 2 \sin \frac{C}{2} \cos \frac{C}{2}$,we get:$ab \sin \frac{C}{2} \cos \frac{C}{2} = \frac{1}{2} l (a + b) \sin \frac{C}{2}$$ab \cos \frac{C}{2} = \frac{1}{2} l (a + b)$Given $\cos \frac{C}{2} = \frac{1}{3}$ and $l = 6$:$ab \left( \frac{1}{3} ight) = \frac{1}{2} (6) (a + b)$$\frac{ab}{3} = 3(a + b)$$\frac{a + b}{ab} = \frac{1}{9}$$\frac{1}{b} + \frac{1}{a} = \frac{1}{9}$

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

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