1.

In the given figure, BOD is the diameter of the circle with centre O. ∠COD= 92° and ∠ABD= 65°. Then y equals(a) 65° (b) 46° (c) 44° (d) 21°

Answer»

(d) 21°

∠BAD = 90° (Angle in a semi-circle) 

∠CAD = \(\frac{1}{2}\)  x ∠COD =(Angle at the centre = 2× angle at any other point on the remaining part of the circle) 

∴ ∠BAC = ∠BAD – ∠CAD = 90°– 46° = 44° 

In ΔABE, ∠AEB = 180° – (65° + 44°) 

= 180° – 109° = 71° 

⇒ ∠CEO = ∠AEB = 71° (Vert. opp. ∠s) 

Also, ∠COE = 180° – 92° = 88° (BOD is a st. line)

∴ In ΔCEO, y = ∠ECO = 180° – (88° + 71°) 

= 180° – 159° = 21°.



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