The document RD Sharma Solutions -Ex-16.4 (Part - 1), Circles, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**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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