Q. 44.0( 14 Votes )

If O is th

Answer :

(i) AOC = 135o

Therefore,


AOC + BOC = 180o (Linear pair)


135o + BOC = 180o


BOC = 45o


By degree measure theorem,


BOC = 2 COB


45o = 2x


x = 22 o


(ii) We have,


ABC = 40o


ACB = 90o (Angle in semi-circle)


In triangle ABC, by angle sum property


CAB + ACB + ABC = 180o


CAB + 90o + 40o = 180o


CAB = 50o


Now,


COB = CAB (Angle on same segment)


x = 50o


(iii) We have,


AOC = 120o


BY degree measure theorem,


AOC = 2 APC


120o= 2 APC


APC = 60o


Therefore,


APC + ABC = 180o (Opposite angles of cyclic quadrilateral)


60o + ABC = 180o


ABC = 120o


ABC + DBC = 180o (Linear pair)


120o + x = 180o


x = 60o


(iv) We have,


CBD = 65o


Therefore,


ABC + CBD = 180o (Linear pair)


ABC + 65o = 1800


ABC = 115o


Therefore,


Reflex AOC = 2 ABC (By degree measure theorem)


x = 2 * 115o


= 230o


(v) We have,


OAB = 35o


Then,


OBA = OAB = 35o (Angle opposite to equal sides are equal)


In triangle AOB, by angle sum property


AOB + OAB + OBA = 180o


AOB + 35o + 35o =180o


AOB = 110o


Therefore,


AOB + Reflex AOB = 360o (Complete angle)


110o + Reflex AOB = 360o


Reflex AOB = 250o


By degree measure theorem,


Reflex AOB = 2 ACB


250o= 2x


x = 125o


(vi) We have,


AOB = 60o


By degree measure theorem,


AOB = 2 ACB


60o = 2 ACB


ACB = 30o


x = 30o


(vii) We have,


BAC = 50o


DBC = 70o


Therefore,


BDC = BAC = 50o (Angles on same segment)


In triangle BDC, by angle sum property


BDC + BCD + DBC = 180o


50o + x + 70o = 180o


120o + x = 180o


x = 60o


(viii) We have,


DBO = 40o


DBC = 90o (Angle in semi-circle)


Therefore,


DBO + OBC = 90o


40o + OBC = 90o


OBC = 50o


By degree measure theorem,


AOC = 2 OBC


x = 2 * 50o


x = 100o


(ix) In triangle DAB, by angle sum property


ADB + DAB + ABD = 180o


32o + DAB + 50o= 180o


DAB = 98o


Now,


DAB + DCB = 180o (Opposite angle of cyclic quadrilateral)


98o + x = 180o


x = 180o – 98o


= 82o


(x) We have,


BAC = 35o


BDC = BAC = 35o (Angle on same segment)


In triangle BCD, by angle sum property


BDC + BCD + DBC = 180o


30o + x + 65o = 180o


x = 80o


(xi) We have,


ABD = 40o


ACD = ABD = 40o (Angle on same segment)


In triangle PCD, by angle sum property


PCD + CPD + PDC = 180o


40o + 110o + x = 180o


x = 30o


(xii) Given that,


BAC = 52o


BDC = BAC = 52o (Angle on same segment)


Since, OD = OC


Then,


ODC = OCD (Opposite angles to equal radii)


x = 52o


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