In a - ABC- the internal angle bisector of - ABC meets AC at K- If BC-2- CK-1 and BK-3-22- then the length of side AB is-

Using cosine law in nbsp;Δ BKC, nbsp; cosB2=4+184 12·3√(2)2·2 ⇒cosB2=3+926√(2)=1512√(2) =54√(2) We know, internal angle bisector, BK=2aca+ccosB2=2·2·2m2m+2 ...

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Byju's Answer

Standard VII

Mathematics

Incentre and Its Construction

In a Δ ABC, t...

Question

In a $Δ A B C$ , the internal angle bisector of $∠ A B C$ meets $A C$ at $K$ . If $B C = 2, C K = 1$ and $B K = \frac{3 \sqrt{2}}{2}$ , then the length of side $A B$ is:

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Solution

Using cosine law in $Δ B K C$ ,
$cos \frac{B}{2} = \frac{4 + \frac{18}{4} - 1}{2 \cdot \frac{3 \sqrt{2}}{2} \cdot 2}$
$\Rightarrow cos \frac{B}{2} = \frac{3 + \frac{9}{2}}{6 \sqrt{2}} = \frac{15}{12 \sqrt{2}} = \frac{5}{4 \sqrt{2}}$
We know, internal angle bisector, $B K = \frac{2 a c}{a + c} cos \frac{B}{2} = \frac{2 \cdot 2 \cdot 2 m}{2 m + 2} \cdot \frac{5}{4 \sqrt{2}}$
$\Rightarrow \frac{3 \sqrt{2}}{2} = \frac{4 m}{m + 1} \cdot \frac{5}{4 \sqrt{2}}$
$\Rightarrow 6 (m + 1) = 10 m \Rightarrow m = \frac{3}{2}$
$∴ A B = 2 m = 3$

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In a ΔABC, the internal angle bisector of ∠ABC meets AC at K. If BC=2,CK=1 and BK=3√22, then the length of side AB is:

Using cosine law in ΔBKC, cosB2=4+184−12⋅3√22⋅2 ⇒cosB2=3+926√2=1512√2=54√2 We know, internal angle bisector, BK=2aca+ccosB2=2⋅2⋅2m2m+2⋅54√2 ⇒3√22=4mm+1⋅54√2 ⇒6(m+1)=10m⇒m=32 ∴AB=2m=3

In a $Δ A B C$ , the internal angle bisector of $∠ A B C$ meets $A C$ at $K$ . If $B C = 2, C K = 1$ and $B K = \frac{3 \sqrt{2}}{2}$ , then the length of side $A B$ is: