Q1) In figure 16.120, O is the centre of the circle. If ∠APB = 500,find ∠AOB and ∠OAB.
Solution:
∠APB = 50^{0}
by degree measure theorem
∠AOB = 2 ∠APB
⇒ ∠APB = 2×50^{0} = 100^{0}
since OA = OB [Radius of circle]
Then ∠OAB = ∠OBA [ Angles opposite to equalsides]
Let ∠OAB = x In ΔOAB, by Angles umproperty ∠OAB+ ∠OBA+ ∠AOB = 180^{0}
= >x + x + 100^{0} = 180^{0}
= >2x = 180^{0} – 100^{0}
= >2x = 80^{0}
= >x = 40^{0}
∠OAB = ∠OBA = 40^{0}
Q2) In figure 16.121, it is given that O is the centre of the circle and ∠AOC = 150^{0}.Find ∠ABC.
Solution:
∠AOC = 150^{0}
∴ ∠AOC+reflex ∠AOC = 360^{0} [Complexangle]
⇒ 150^{0}+reflex ∠AOC = 360^{0}
⇒ reflex ∠AOC = 360^{0}−150^{0}
⇒ reflex ∠AOC = 210^{0}
⇒ 2 ∠ABC = 210^{0} [ by degree measure theorem]
⇒ ∠ABC = = 105^{0}
Q3) In figure 16.22, O is the centre of the circle. Find ∠BAC.
Solution:
We have ∠AOB = 80^{0}
and ∠AOC = 110^{0}
Therefore, ∠AOB+ ∠AOC+ ∠BOC = 360^{0} [Completeangle]
⇒ 80^{0}+100^{0}+ ∠BOC = 360^{0}
⇒ ∠BOC = 360^{0}−80^{0}−110^{0}
⇒ ∠BOC = 170^{0}
by degree measure theorem
∠BOC = 2 ∠BAC
⇒ 170^{0} = 2 ∠BAC
⇒ ∠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:
∠AOC = 135^{0}
∴ ∠AOC+ ∠BOC = 180^{0} [Linearpair of Angles ]
⇒ 135^{0 }+ ∠BOC = 180^{0}
⇒ ∠BOC = 180^{0}−135^{0}
⇒ ∠BOC = 45^{0}
by degree measure theorem ∠BOC = 2 ∠CPB
⇒ 45^{0} = 2x
(ii)
Solution:
Wehave ∠ABC = 40^{0} ∠ACB = 90^{0} [Angle in semicircle]
In ΔABC, by Angles umproperty ∠CAB+ ∠ACB+ ∠ABC = 180^{0}
⇒ ∠CAB+90^{0}+40^{0 }= 180^{0}
⇒ ∠CAB = 180^{0}−90^{0}−40^{0}
⇒ ∠CAB = 50^{0}
Now, ∠CDB = ∠CAB [Angleissameinsegment]
⇒ x = 50^{0}
(iii)
Solution:
Wehave ∠AOC = 120^{0 }by degree measure theorem. ∠AOC = 2 ∠APC
⇒ 120^{0 }= 2 ∠APC
⇒ ∠APC = = 60^{0}
∠APC+ ∠ABC = 180^{0} [Opposite Angles of cyclicquadrilaterals]
⇒ 60^{0}+ ∠ABC = 180^{0}
⇒ ∠ABC = 180^{0}−60^{0}
⇒ ∠ABC = 120^{0}
∴ ∠ABC+ ∠DBC = 180^{0} [Linearpair of Angles ]
⇒ 12^{0}+x = 180^{0}
⇒ x = 180^{0}−120^{0} = 60^{0}
(iv)
Solution:
Wehave ∠CBD = 65^{0}
∴ ∠ABC+ ∠CBD = 180^{0} [Linearpair of Angles ]
⇒ ∠ABC = 65^{0} = 180^{0}
⇒ ∠ABC = 180^{0}−65^{0 }= 115^{0}
∴reflex ∠AOC = 2 ∠ABC [ by degree measure theorem]
⇒ x = 2×115^{0}
⇒ x = 230^{0}
(v)
Solution:
Wehave ∠OAB = 35^{0}
Then, ∠OBA = ∠OAB = 35^{0} [ Angles opposite to equalradii]
InΔAOB, by Angles umproperty
⇒ ∠AOB+ ∠OAB+ ∠OBA = 180^{0}
⇒ ∠AOB+35^{0}+35^{0} = 180^{0}
⇒ ∠AOB = 180^{0}−35^{0}−35^{0} = 110^{0}
∴ ∠AOB+reflex ∠AOB = 360^{0} [Complexangle]
⇒ 110^{0}+reflex ∠AOB = 360^{0}
⇒ reflex ∠AOB = 360^{0}−110^{0} = 250^{0} by degree measure theoremreflex ∠AOB = 2 ∠ACB
⇒ 250^{0} = 2x
⇒ x = = 125^{0}
(vi)
Solution:
Wehave ∠AOB = 60^{0} by degree measure theoremreflex ∠AOB = 2 ∠ACB
⇒ 60^{0 }= 2 ∠ACB
⇒ ∠ACB = = 30^{0} [ Angles opposite to equalradii]
⇒ x = 30^{0}.
(vii)
Solution:
Wehave ∠BAC = 50^{0} and ∠DBC = 70^{0}
∴ ∠BDC = ∠BAC = 50^{0} [Angleinsamesegment]
InΔBDC, by Angles umproperty ∠BDC+ ∠BCD+ ∠DBC = 180^{0}
⇒ 50^{0}+x+70^{0 }= 180^{0}
⇒ x = 180^{0}−50^{0}−70^{0} = 60^{0}
(viii)
Solution:
Wehave, ∠DBO = 40^{0} and ∠DBC = 90^{0} [Angleinasemicircle]
⇒ ∠DBO+ ∠OBC = 90^{0}
⇒ 40^{0}+ ∠OBC = 90^{0}
⇒ ∠OBC = 90^{0}−40^{0 }= 50^{0} by degree measure theorem
∠AOC = 2 ∠OBC
⇒ x = 2×50^{0} = 100^{0}
(ix)
Solution:
InΔDAB, by Angles umproperty ∠ADB+ ∠DAB+ ∠ABD = 180^{0}
⇒ 32^{0}+ ∠DAB+50^{0} = 180^{0}
⇒ ∠DAB = 180^{0}−32^{0}−50^{0}
⇒ ∠DAB = 98^{0}
Now, ∠OAB+ ∠DCB = 180^{0} [Opposite Angles of cyclicquadrilateral]
⇒ 98^{0 }+ x = 180^{0}
⇒ x = 180^{0}−98^{0} = 82^{0}
(x)
Solution:
Wehave, ∠BAC = 35^{0}
∠BDC = ∠BAC = 35^{0} [Angleinsamesegment]
InΔBCD, by Angles umproperty ∠BDC+ ∠BCD+ ∠DBC = 180^{0}
⇒ 35^{0}+x+65^{0} = 180^{0}
⇒ x = 180^{0}−35^{0}−65^{0} = 80^{0}
(xi)
Solution:
Wehave, ∠ABD = 40^{0}
∠ACD = ∠ABD = 40^{0} [Angleinsamesegment]
InΔPCD, by Angles umproperty
∠PCD+ ∠CPO+ ∠PDC = 180^{0}
⇒ 40^{0}+110^{0}+x = 180^{0}
⇒ x = 180^{0}−150^{0}
⇒ x = 30^{0}
(xii)
Solution:
Giventhat, ∠BAC = 52^{0}
Then ∠BDC = ∠BAC = 52^{0} [Angleinsamesegment]
SinceOD = OC
Then ∠ODC = ∠OCD [Oppositeangle to equalradii]
⇒ x = 52^{0}
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