Hi, In triangle OBA , BC is bisector of angle OBA so OBAB= OCACABOB= ACOCadding 1 both side AB+OBOB= AC+OC OC= OAOC= 1OCSo OC = OBAB+OB 1now in right angle triangle BAO, sinθ= ABOBSO AB = OB sinθputting in first equation, OC = OBOB sinθ+OB=11+sinθ Regards

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Hi,

In triangle OBA, BC is bisector of angle OBA so \frac{OB}{AB} = \frac{OC}{AC} \frac{AB}{OB} = \frac{AC}{OC} adding 1 both side \frac{AB + OB}{OB} = \frac{AC + OC}{OC} = \frac{OA}{OC} = \frac{1}{OC} So OC = \frac{OB}{AB + OB} . . . . . (1) now in right angle triangle BAO, sinθ = \frac{AB}{OB} SO AB = OB sinθ putting in first equation, OC = \frac{OB}{OB sinθ + OB} = \frac{1}{1 + sinθ}

Regards

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