1.

If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that angleDBC=120^(@), prove that BC+BD=BO.

Answer»

SOLUTION :`angle1+angle2=120^(@)""`(fiven)
But `angle1=angle2""`(tangents are equally inclined at the centre)
`:.""angle1+angle1=120^(@)`
`implies""2angle1=120^(@)`
`implies""angle1=60^(@)`
ALSO, `angleOCB=90^(@)`
(radius through point of contact is `_|_` to the tangent)
Now, in right `triangleOCB,`
`cos60^(@)=(BC)/(OB)`
CIRCLES
`implies""(1)/(2)=(BC)/(OB)`
`:.""OB=2BC""implies""OB=BC+BC`
`implies""OB-BC+BD`(`:.` length of tangents from an external point are equal)

Hence Proved.


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