Question 9. In the fig.Angle DBC =145 degree.Find angle BAC
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∠OBD=90°(tangent-radius property)
∠DBC=∠OBD+∠OBC⇒145°=90°+∠OBC⇒∠OBC=145-90=55°
Also,OB=OC(both are radii of same circle)⇒∠OBC=∠OCB=55°
In ΔBOC,∠BOC+∠OBC+∠OCB=180°⇒∠BOC+55°+55°=180°
∠BOC+110°=180°⇒∠BOC=180-110=70°
∠BOC=2∠BAC (arc-centre property)
∠BAC=1/2(∠BOC)=1/2(70)=35°
∴∠BAC=35°
∠DBC=∠OBD+∠OBC⇒145°=90°+∠OBC⇒∠OBC=145-90=55°
Also,OB=OC(both are radii of same circle)⇒∠OBC=∠OCB=55°
In ΔBOC,∠BOC+∠OBC+∠OCB=180°⇒∠BOC+55°+55°=180°
∠BOC+110°=180°⇒∠BOC=180-110=70°
∠BOC=2∠BAC (arc-centre property)
∠BAC=1/2(∠BOC)=1/2(70)=35°
∴∠BAC=35°
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