If in a ABC- CD is the angle bisector of the angle ACB- then CD- kab - a-b cos C - 2

The correct option is C 2We have, in Δ ADC #160; CD /sin A #160; = AD /sin C 2 #160; #160;⇒ AD=CDsin C 2 #160; #160;/sin B #160; Simi ...

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Byju's Answer

Standard XII

Mathematics

Property 2

If in a ABC...

Question

If in a $△ A B C$ , $C D$ is the angle bisector of the angle $A C B$ , then $C D = \frac{k a b}{a + b} cos \frac{C}{2}$

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Solution

The correct option is C 2
We have, in $Δ A D C \frac{C D}{sin A} = \frac{A D}{sin \frac{C}{2}} \Rightarrow A D = C D \frac{sin \frac{C}{2}}{sin B}$

Similarly, in
$Δ B D C, B D = C D \frac{sin \frac{C}{2}}{sin B}$

Now, $A D + B D = c \Rightarrow C D . sin \frac{C}{2} (\frac{1}{sin A} + \frac{1}{sin B}) = c$

$\Rightarrow C D = c \frac{sin A . sin B}{(sin A + sin B) sin \frac{C}{2}} = 2 c \frac{sin A . sin B . cos \frac{C}{2}}{(sin A + sin B) sin C}$

$= \frac{2 c a . b cos \frac{C}{2}}{(a + b) c} = \frac{2 a b}{a + b} cos \frac{C}{2}$ [by sine rule]

by comparing it with $\frac{k a b}{a + b} c o s \frac{C}{2}$ we get

$\Rightarrow k = 2$

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If in a △ABC, CD is the angle bisector of the angle ACB, then CD=kaba+bcosC2

The correct option is C 2We have, in ΔADCCDsinA=ADsinC2⇒AD=CDsinC2sinBSimilarly, in ΔBDC,BD=CDsinC2sinBNow, AD+BD=c⇒CD.sinC2(1sinA+1sinB)=c⇒CD=csinA.sinB(sinA+sinB)sinC2=2csinA.sinB.cosC2(sinA+sinB)sinC=2ca.bcosC2(a+b)c=2aba+bcosC2 [by sine rule]by comparing it with kaba+bcosC2 we get⇒k=2

If in a $△ A B C$ , $C D$ is the angle bisector of the angle $A C B$ , then $C D = \frac{k a b}{a + b} cos \frac{C}{2}$