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In the isosceles triangle ABC, |vec(AB)...

In the isosceles triangle `ABC, |vec(AB)| = |vec(BC)| = 8`,a point E divide AB internally in the ratio `1:3`, then the cosine of the angle between `vec(CE)` and `vec(CA)` is (where `|vec(CA)| = 12`)

Text Solution

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`abs(AB)=abs(BC)=8`
`abs(vecb)=8`
`abs(vecc)=12`
`abs(vecb-vecc)=8`
`vec(CE)=vecE-vecC`
`=vecb/4-vecC`
Therefore, `abs(vecCE)=sqrt(148-(vecb*vecc)/2)`
`a/sina=b/sinb=c/sinc`
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