The RD Sharma Class 9 solutions for the chapter “Circle” exercise 16.4 is given here. In this exercise, students will learn about arcs and angles subtended by them. All the answers solved by subject experts using step-by-step problem solving approach which can help students understand this topic in a better and fun way.

## Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.4

### Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 16 Circles Exercise 16.4 Page number 16.60

**Question 1: In figure, O is the centre of the circle. If ∠APB = 50 ^{0}, find ∠AOB and ∠OAB.**

**Solution:**

∠APB = 50^{0} (Given)

By degree measure theorem: ∠AOB = 2∠APB

∠AOB = 2 × 50^{0} = 100^{0}

Again, OA = OB [Radius of circle]

Then ∠OAB = ∠OBA [Angles opposite to equal sides]

Let ∠OAB = m

In ΔOAB,

By angle sum property: ∠OAB+∠OBA+∠AOB=180^{0}

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

=>2m = 180^{0} – 100^{0} = 80^{0}

=>m = 80^{0}/2 = 40^{0}

∠OAB = ∠OBA = 40^{0}

**Question 2: In figure, it is given that O is the centre of the circle and ∠AOC = 150 ^{0}. Find ∠ABC.**

**Solution:**

∠AOC = 150^{0} (Given)

By degree measure theorem: ∠ABC = (reflex∠AOC)/2 …(1)

We know, ∠AOC + reflex(∠AOC) = 360^{0} [Complex angle]

150^{0} + reflex∠AOC = 360^{0}

or reflex ∠AOC = 360^{0}−150^{0} = 210^{0}

From (1) => ∠ABC = 210^{ o} /2 = 105^{o}

**Question 3: In figure, O is the centre of the circle. Find ∠BAC. **

** **

** Solution:**

Given: ∠AOB = 80^{0} and ∠AOC = 110^{0}

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

Substitute given values,

80^{0} + 100^{0} + ∠BOC = 360^{0}

∠BOC = 360^{0} – 80^{0} – 110^{0} = 170^{0}

or ∠BOC = 170^{0}

Now, by degree measure theorem

∠BOC = 2∠BAC

170^{0} = 2∠BAC

Or ∠BAC = 170^{0}/2 = 85^{0}

**Question 4: If O is the centre of the circle, find the value of x in each of the following figures.**

**(i)**

**Solution:**

∠AOC = 135^{0} (Given)

From figure, ∠AOC + ∠BOC = 180^{0} [Linear pair of angles]

135^{0} +∠BOC = 180^{0}

or ∠BOC=180^{0}−135^{0}

or ∠BOC=45^{0}

Again, by degree measure theorem

∠BOC = 2∠CPB

45^{0} = 2x

x = 45^{0}/2

**(ii)**

**Solution:**

∠ABC=40^{0} (given)

∠ACB = 90^{0} [Angle in semicircle]

In ΔABC,

∠CAB+∠ACB+∠ABC=180^{0} [angle sum property]

∠CAB+90^{0}+40^{0}=180^{0}

∠CAB=180^{0}−90^{0}−40^{0}

∠CAB=50^{0}

Now, ∠CDB = ∠CAB [Angle is on same segment]

This implies, x = 50^{0}

**(iii)**

**Solution:**

∠AOC = 120^{0} (given)

By degree measure theorem: ∠AOC = 2∠APC

120^{0 }= 2∠APC

∠APC = 120^{0}/2 = 60^{0}

Again, ∠APC + ∠ABC = 180^{0} [Sum of opposite angles of cyclic quadrilaterals = 180 ^{o} ]

60^{0} + ∠ABC=180^{0}

∠ABC=180^{0}−60^{0}

∠ABC = 120^{0}

∠ABC + ∠DBC = 180^{0} [Linear pair of angles]

120^{0 }+ x = 180^{0}

x = 180^{0}−120^{0}=60^{0}

The value of x is 60^{0}

**(iv) **

**Solution:**

∠CBD = 65^{0} (given)

From figure:

∠ABC + ∠CBD = 180^{0} [ Linear pair of angles]

∠ABC + 65^{0} = 180^{0}

∠ABC =180^{0}−65^{0}=115^{0}

Again, reflex ∠AOC = 2∠ABC [Degree measure theorem]

x=2(115^{0}) = 230^{0}

The value of x is 230^{0}

**(v)**

**Solution:**

∠OAB = 35^{0} (Given)

From figure:

∠OBA = ∠OAB = 35^{0} [Angles opposite to equal radii]

InΔAOB:

∠AOB + ∠OAB + ∠OBA = 180^{0} [angle sum property]

∠AOB + 35^{0} + 35^{0} = 180^{0}

∠AOB = 180^{0} – 35^{0} – 35^{0} = 110^{0}

Now, ∠AOB + reflex∠AOB = 360^{0} [Complex angle]

110^{0} + reflex∠AOB = 360^{0}

reflex∠AOB = 360^{0} – 110^{0} = 250^{0}

By degree measure theorem: reflex ∠AOB = 2∠ACB

250^{0} = 2x

x = 250^{0}/2=125^{0}

**(vi)**

**Solution:**

∠AOB = 60^{o} (given)

By degree measure theorem: reflex∠AOB = 2∠OAC

60^{ o} = 2∠ OAC

∠OAC = 60^{ o} / 2 = 30^{ o} [Angles opposite to equal radii]

Or x = 30^{0}

**(vii)**

**Solution:**

∠BAC = 50^{0} and ∠DBC = 70^{0} (given)

From figure:

∠BDC = ∠BAC = 50^{0 }[Angle on same segment]

Now,

In ΔBDC:

Using angle sum property, we have

∠BDC+∠BCD+∠DBC=180^{0}

Substituting given values, we get

50^{0} + x^{0} + 70^{0} = 180^{0}

x^{0 }= 180^{0}−50^{0}−70^{0}=60^{0}

or x = 60^{o}

**(viii) **

**Solution:**

∠DBO = 40^{0} (Given)

Form figure:

∠DBC = 90^{0} [Angle in a semicircle]

∠DBO + ∠OBC = 90^{0}

40^{0}+∠OBC=90^{0}

or ∠OBC=90^{0}−40^{0}=50^{0}

Again, By degree measure theorem: ∠AOC = 2∠OBC

or x = 2×50^{0}=100^{0}

**(ix)**

**Solution:**

∠CAD = 28, ∠ADB = 32 and ∠ABC = 50 (Given)

From figure:

In ΔDAB:

Angle sum property: ∠ADB + ∠DAB + ∠ABD = 180^{0}

By substituting the given values, we get

32^{0} + ∠DAB + 50^{0} = 180^{0}

∠DAB=180^{0}−32^{0}−50^{0}

∠DAB = 98^{0}

Now,

∠DAB+∠DCB=180^{0 } [Opposite angles of cyclic quadrilateral, their sum = 180 degrees]

98^{0}+x=180^{0}

or x = 180^{0}−98^{0}=82^{0}

The value of x is 82 degrees.

**(x)**

**Solution:**

∠BAC = 35^{0} and ∠DBC = 65^{0}

From figure:

∠BDC = ∠BAC = 35^{0} [Angle in same segment]

In ΔBCD:

Angle sum property, we have

∠BDC + ∠BCD + ∠DBC = 180^{0}

35^{0} + x + 65^{0} = 180^{0}

or x = 180^{0} – 35^{0} – 65^{0} = 80^{0}

**(xi)**

**Solution:**

∠ABD = 40^{0}, ∠CPD = 110^{0} (Given)

Form figure:

∠ACD = ∠ABD = 40^{0} [Angle in same segment]

In ΔPCD,

Angle sum property: ∠PCD+∠CPO+∠PDC=180^{0}

400 + 110^{0} + x = 180^{0}

x=180^{0}−150^{0 }=30^{0}

The value of x is 30 degrees.

**(xii)**

**Solution:**

∠BAC = 52^{0} (Given)

From figure:

∠BDC = ∠BAC = 52^{0} [Angle in same segment]

Since OD = OC (radii), then ∠ODC = ∠OCD [Opposite angle to equal radii]

So, x = 52^{0}

**Question 5: O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A.**

**Solution:**

In ΔOBD and ΔOCD:

OB = OC [Radius]

∠ODB = ∠ODC [Each 90^{0}]

OD = OD [Common]

Therefore, By RHS Condition

ΔOBD ≅ ΔOCD

So, ∠BOD = ∠COD…..(i)[By CPCT]

Again,

By degree measure theorem: ∠BOC = 2∠BAC

2∠BOD = 2∠BAC [Using(i)]

∠BOD = ∠BAC

Hence proved.

**Question 6: In figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC. **

** **

**Solution:**

Since, BO is the bisector of ∠ABC, then,

∠ABO = ∠CBO …..(i)

From figure:

Radius of circle = OB = OA = OB = OC

∠OAB = ∠OCB …..(ii) [opposite angles to equal sides]

∠ABO = ∠DAB …..(iii) [opposite angles to equal sides]

From equations (i), (ii) and (iii), we get

∠OAB = ∠OCB …..(iv)

In ΔOAB and ΔOCB:

∠OAB = ∠OCB [From (iv)]

OB = OB [Common]

∠OBA = ∠OBC [Given]

Then, By AAS condition : ΔOAB ≅ ΔOCB

So, AB = BC [By CPCT]

**Question 7: In figure, O is the centre of the circle, then prove that ∠x = ∠y + ∠z.**

**Solution: **

From the figure:

∠3 = ∠4 ….(i) [Angles in same segment]

∠x = 2∠3 [By degree measure theorem]

∠x = ∠3 + ∠3

∠x = ∠3 + ∠4 (Using (i) ) …..(ii)

Again, ∠y = ∠3 + ∠1 [By exterior angle property]

or ∠3 = ∠y − ∠1 …..(iii)

∠4 = ∠z + ∠1 …. (iv) [By exterior angle property]

Now, from equations (ii) , (iii) and (iv), we get

∠x = ∠y − ∠1 + ∠z + ∠1

or ∠x = ∠y + ∠z + ∠1 − ∠1

or x = ∠y + ∠z

Hence proved.

## RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.4

Class 9 Maths Chapter 16 Circles Exercise 16.4 is based on the topic – Arcs and angles subtended by them. Students can practice more Maths concepts using RD Sharma solutions and clear their doubts.