Q31 of 63 Page 156

In figure, O is the centre of a circle and ∠BCO = 30°. Find x and y.

Given O is the centre of the circle and ∠BCO = 30°.


Construction:


Join AC and OB.



In ΔOBC,


⇒ OC = OB [Radii of circle]


⇒∠OBC = ∠OCB = 30°


From angle sum property,


⇒∠OBC + ∠OCB + ∠BOC = 180°


⇒ 30° +30° + ∠BOC = 180°


⇒∠BOC = 180° - 60°


∴ ∠BOC = 120°


We know that the central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.


⇒∠BOC = 2∠BAC


⇒ 2∠BAC = 120°


∴ ∠BAC = 60°


⇒∠AEB = 90°


So, OE ⊥ BC


We know that a line form center to any chord is perpendicular then that line also bisects chord.


⇒ CE = BE … (1)


Consider ΔABE and ΔACE,


⇒ CE = BE [From (1)]


⇒∠AEB = ∠AEC = 90°


⇒ AE = AE [Common side]


By RHL rule,


⇒ ΔABE ≅ ΔACE


By CPCT,


∠BAE = ∠CAE


⇒∠BAC = ∠BAE + ∠CAE


⇒ x + x = 60°


⇒ 2x = 60°


∴ x = 30°


And ∠COI = ∠OCB = 30° [Alternate interior angles]


We know that the central angle subtended by two points on a circle is twice the inscribed angle subtended by those points.


⇒∠COI = 2∠CBI


⇒ 2y = 30°


∴ y = 15°


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