KVPY 2020-21 SB SX Exam Previous Papers

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Question : 19 of 120

Marks: +1, -0

Let

{ABC}

be a triangle such that

{AB}=15

and

{AC}=9

. The bisector of

\∠{BAC}

meets

{BC}

in D. If

\∠{ACB}=2 \∠{ABC}

, then

{BD}

is

8
9
10
12

Solution:

In ΔABC

{15}/{\sin 2 \θ}={9}/{\sin \θ}={{BC}}/{\sin 3 \θ}

{15}/{\sin 2 \θ}={9}/{\sin \θ} \⇒\cos \θ={5}/{6}

{9}/{\sin \θ}={{BC}}/{\sin 3 \θ}

\⇒{BC}=9[3-4 \sin ^{2} \θ]

=9[4 \cos ^{2} \θ-1]

=9[4 \× {25}/{36}-1]

=16

\∴B D={5}/{8} B C=10

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