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**Q1) In figure 16.120, O is the centre of the circle. If âˆ APB = 500,find âˆ AOB and âˆ OAB.**

**Solution:**

âˆ APB = 50^{0}

by degree measure theorem

âˆ AOB = 2 âˆ APB

â‡’ âˆ APB = 2Ã—50^{0} = 100^{0}

since OA = OB [Radius of circle]

Then âˆ OAB = âˆ OBA [ Angles opposite to equalsides]

Let âˆ OAB = x In Î”OAB, by Angles umproperty âˆ OAB+ âˆ OBA+ âˆ AOB = 180^{0}

= >x + x + 100^{0} = 180^{0}

= >2x = 180^{0} â€“ 100^{0}

= >2x = 80^{0}

= >x = 40^{0}

âˆ OAB = âˆ OBA = 40^{0}

**Q2) In figure 16.121, it is given that O is the centre of the circle and âˆ AOC = 150 ^{0}.Find âˆ ABC.**

**Solution:**

âˆ AOC = 150^{0}

âˆ´ âˆ AOC+reflex âˆ AOC = 360^{0} [Complexangle]

â‡’ 150^{0}+reflex âˆ AOC = 360^{0}

â‡’ reflex âˆ AOC = 360^{0}âˆ’150^{0}

â‡’ reflex âˆ AOC = 210^{0}

â‡’ 2 âˆ ABC = 210^{0} [ by degree measure theorem]

â‡’ âˆ ABC = = 105^{0}

** **** **

**Q3) In figure 16.22, O is the centre of the circle. Find âˆ BAC.**

** ****Solution:**

We have âˆ AOB = 80^{0}

and âˆ AOC = 110^{0}

Therefore, âˆ AOB+ âˆ AOC+ âˆ BOC = 360^{0} [Completeangle]

â‡’ 80^{0}+100^{0}+ âˆ BOC = 360^{0}

â‡’ âˆ BOC = 360^{0}âˆ’80^{0}âˆ’110^{0}

â‡’ âˆ BOC = 170^{0}

by degree measure theorem

âˆ BOC = 2 âˆ BAC

â‡’ 170^{0} = 2 âˆ BAC

â‡’ âˆ BAC = = 85^{0}

**Q4) If O is the centre of the circle, find the value of x in each of the following figures.**

**(i)**

**Solution:**

âˆ AOC = 135^{0}

âˆ´ âˆ AOC+ âˆ BOC = 180^{0} [Linearpair of Angles ]

â‡’ 135^{0 }+ âˆ BOC = 180^{0}

â‡’ âˆ BOC = 180^{0}âˆ’135^{0}

â‡’ âˆ BOC = 45^{0}

by degree measure theorem âˆ BOC = 2 âˆ CPB

â‡’ 45^{0} = 2x

**(ii)**

**Solution:**

Wehave âˆ ABC = 40^{0} âˆ ACB = 90^{0} [Angle in semicircle]

In Î”ABC, by Angles umproperty âˆ CAB+ âˆ ACB+ âˆ ABC = 180^{0}

â‡’ âˆ CAB+90^{0}+40^{0 }= 180^{0}

â‡’ âˆ CAB = 180^{0}âˆ’90^{0}âˆ’40^{0}

â‡’ âˆ CAB = 50^{0}

Now, âˆ CDB = âˆ CAB [Angleissameinsegment]

â‡’ x = 50^{0}

**(iii)**

**Solution:**

Wehave âˆ AOC = 120^{0 }by degree measure theorem. âˆ AOC = 2 âˆ APC

â‡’ 120^{0 }= 2 âˆ APC

â‡’ âˆ APC = = 60^{0}

âˆ APC+ âˆ ABC = 180^{0} [Opposite Angles of cyclicquadrilaterals]

â‡’ 60^{0}+ âˆ ABC = 180^{0}

â‡’ âˆ ABC = 180^{0}âˆ’60^{0}

â‡’ âˆ ABC = 120^{0}

âˆ´ âˆ ABC+ âˆ DBC = 180^{0} [Linearpair of Angles ]

â‡’ 12^{0}+x = 180^{0}

â‡’ x = 180^{0}âˆ’120^{0} = 60^{0}

**(iv) **

**Solution:**

Wehave âˆ CBD = 65^{0}

âˆ´ âˆ ABC+ âˆ CBD = 180^{0} [Linearpair of Angles ]

â‡’ âˆ ABC = 65^{0} = 180^{0}

â‡’ âˆ ABC = 180^{0}âˆ’65^{0 }= 115^{0}

âˆ´reflex âˆ AOC = 2 âˆ ABC [ by degree measure theorem]

â‡’ x = 2Ã—115^{0}

â‡’ x = 230^{0}

**(v)**

**Solution:**

Wehave âˆ OAB = 35^{0}

Then, âˆ OBA = âˆ OAB = 35^{0} [ Angles opposite to equalradii]

InÎ”AOB, by Angles umproperty

â‡’ âˆ AOB+ âˆ OAB+ âˆ OBA = 180^{0}

â‡’ âˆ AOB+35^{0}+35^{0} = 180^{0}

â‡’ âˆ AOB = 180^{0}âˆ’35^{0}âˆ’35^{0} = 110^{0}

âˆ´ âˆ AOB+reflex âˆ AOB = 360^{0} [Complexangle]

â‡’ 110^{0}+reflex âˆ AOB = 360^{0}

â‡’ reflex âˆ AOB = 360^{0}âˆ’110^{0} = 250^{0} by degree measure theoremreflex âˆ AOB = 2 âˆ ACB

â‡’ 250^{0} = 2x

â‡’ x = = 125^{0}

**(vi)**

**Solution:**

Wehave âˆ AOB = 60^{0} by degree measure theoremreflex âˆ AOB = 2 âˆ ACB

â‡’ 60^{0 }= 2 âˆ ACB

â‡’ âˆ ACB = = 30^{0} [ Angles opposite to equalradii]

â‡’ x = 30^{0}.

**(vii)**

**Solution:**

Wehave âˆ BAC = 50^{0} and âˆ DBC = 70^{0}

âˆ´ âˆ BDC = âˆ BAC = 50^{0} [Angleinsamesegment]

InÎ”BDC, by Angles umproperty âˆ BDC+ âˆ BCD+ âˆ DBC = 180^{0}

â‡’ 50^{0}+x+70^{0 }= 180^{0}

â‡’ x = 180^{0}âˆ’50^{0}âˆ’70^{0} = 60^{0}

**(viii) **

**Solution:**

Wehave, âˆ DBO = 40^{0} and âˆ DBC = 90^{0} [Angleinasemicircle]

â‡’ âˆ DBO+ âˆ OBC = 90^{0}

â‡’ 40^{0}+ âˆ OBC = 90^{0}

â‡’ âˆ OBC = 90^{0}âˆ’40^{0 }= 50^{0} by degree measure theorem

âˆ AOC = 2 âˆ OBC

â‡’ x = 2Ã—50^{0} = 100^{0}

**(ix)**

**Solution:**

InÎ”DAB, by Angles umproperty âˆ ADB+ âˆ DAB+ âˆ ABD = 180^{0}

â‡’ 32^{0}+ âˆ DAB+50^{0} = 180^{0}

â‡’ âˆ DAB = 180^{0}âˆ’32^{0}âˆ’50^{0}

â‡’ âˆ DAB = 98^{0}

Now, âˆ OAB+ âˆ DCB = 180^{0} [Opposite Angles of cyclicquadrilateral]

â‡’ 98^{0 }+ x = 180^{0}

â‡’ x = 180^{0}âˆ’98^{0} = 82^{0}

**(x)**

**Solution:**

Wehave, âˆ BAC = 35^{0}

âˆ BDC = âˆ BAC = 35^{0} [Angleinsamesegment]

InÎ”BCD, by Angles umproperty âˆ BDC+ âˆ BCD+ âˆ DBC = 180^{0}

â‡’ 35^{0}+x+65^{0} = 180^{0}

â‡’ x = 180^{0}âˆ’35^{0}âˆ’65^{0} = 80^{0}

**(xi)**

**Solution:**

Wehave, âˆ ABD = 40^{0}

âˆ ACD = âˆ ABD = 40^{0} [Angleinsamesegment]

InÎ”PCD, by Angles umproperty

âˆ PCD+ âˆ CPO+ âˆ PDC = 180^{0}

â‡’ 40^{0}+110^{0}+x = 180^{0}

â‡’ x = 180^{0}âˆ’150^{0}

â‡’ x = 30^{0}

** (xii)**

**Solution:**

Giventhat, âˆ BAC = 52^{0}

Then âˆ BDC = âˆ BAC = 52^{0} [Angleinsamesegment]

SinceOD = OC

Then âˆ ODC = âˆ OCD [Oppositeangle to equalradii]

â‡’ x = 52^{0}