RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.3 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 12 Circle Exercise 12.3.

## Rajasthan Board RBSE Class 10 Maths Chapter 12 Circle Ex 12.3

Question 1.

Write true/false for each (RBSESolutions.com) and give answer with reason.

(i) Angle subtended by the chord at any two points on the circle are equal.

(ii) In figure, if AB is diameter of circle and C is any point on circle, then AC^{2} + BC^{2} = AB^{2}

(iii) In figure, if ∠ADE = 120°, then ∠EAB = 30°

(iv) In fig. ∠CAD = ∠CED

Solution :

(i) **False** – If two points lie in same (RBSESolutions.com) segment they are equal otherwise not.

(ii) **True** – Since angle formed in semicircle is right angle

i.e., ∠C = 90°

∴ AC^{2} + BC^{2} = AB^{2}

(iii) **True** – After joining AD, DE, DB, EB and AE

∠ADB = 90° (Angle in semicircle)

∵ ∠ADE = 120° (given)

∴ ∠BDE = 120° – 90° = 30°

here ∠BDE = ∠EAB = 30° (Since angles are of same segment)

(iv) **True** – Since equal chords of (RBSESolutions.com) congruent triangles subtends equal angles at corresponding centres.

Question 2.

In figure, if ∠ABC 45°, then prove that OA ⊥ OC.

Solution :

∠ABC = 45° (given)

We know that angle subtended by an arc of a (RBSESolutions.com) circle at center is double the angle subtended at remaining part of circle by same arc.

Then ∠AOC = 2∠ABC

= 2 × 45°

= 90°

Thus, OA ⊥ OC

Question 3.

If O is circumcenter of ∆ABC and D is mid point of base BC, then prove that ∠BOD = ∠A.

Solution :

Given :

∆ABC whose circumcenter is point O and D is mid point of BC

∴ BD = DC

To Proved :

∠BOD = ∠A

Construction: Draw OD ⊥ BC and join OB and OC

In ∆OBD and ∆OCD

OB = OC (radius of (RBSESolutions.com) same circle)

∠ODB = ∠ODC (each 90°)

OD = OD (common)

∴ ∆OBD = ∆OCD

⇒ ∠BOD = ∠COD

⇒ ∠BOC = 2∠BOD = 2∠COD

∵ Angle subtended by arc BC at the center of circle i.e., ∠BOC is double the angle subtended by same arc at point A of remaining part of circle i.e., ∠BAC.

∠BOC = 2∠A

⇒ 2∠BOD = 2∠A [∵∠BOC = 2∠BOD]

⇒ ∠BOD = ∠A

Question 4.

If two triangles ∆ACB and ∆ADB are drawn (RBSESolutions.com) on both sides of common hypotenuse AB, then Prove that ∠BAC = ∠BDC.

Solution :

Taking AB as diameter, draw a circle which passes through points C and B

∵ ∠ACB = ∠ADB

= 90° [Since angle in semicircle is 90°]

Thus, angles subtended by minor arc BC is same segment will be equal.

∴∠BAC = ∠BDC

Question 5.

Two chords AB and AC subtend angle 90° and 150° respectively (RBSESolutions.com) at center of the circle. Find ∠BAC if AB and AC lie on opposite sides of center.

Solution :

AB and AC arc two chords of circle. Which makes an angles 90° and 150° with centre

∠BOA = 90° and ∠AOC = 150°

∴ ∠BOC = 360° – (∠BOA + ∠AOC)

= 360° – (90° + 150°)

= 360° – 240°

= 120°

we know that

∠BAC = \(\frac { 1 }{ 2 }\) ∠BOC

= \(\frac { 1 }{ 2 }\) × 120°

= 60°

thus ∠BAC = 60°

Question 6.

In O is circurncenter of ∆ABC, Prove (RBSESolutions.com) that ∠OBC + ∠BAC = 90°

Solution : O is circumcentre of ∆ABC, join OB and OC. We know that

∴ ∠BOC = 2∠BAC

z = 2x

In ∆OBC

∠OBC + ∠OCB + ∠BOC = 180°

⇒ y + z + y = 180°

⇒ 2y + z = 180°

⇒ 2y + 2x = 180°

⇒ 2(x + y) = 180°

⇒ x + y = 90°

⇒ ∠BAC + ∠OBC = 90°

or ∠OBC + ∠BAC = 90°

Question 7.

A chord and radius of a circle are equal. (RBSESolutions.com) find the angle subtended by this chord in major segment at any point

Solution :

Let AB is chord of circle chord AB = radius OA radius OB

∆AOB is an equilateral triangle

⇒ ∠AOB = 60°

[Each angle of equilateral triangle in 60°]

Let C is any point on major segment BA

∠ACB = \(\frac { 1 }{ 2 }\)∠AOB

= \(\frac { 1 }{ 2 }\) × 60°

= 30°

[Since angle subtended by arc of circle (RBSESolutions.com) at center is double the angle subtended at remaining part of circle]

Thus, angle subtended by chord at any point in major segment = 30°

Question 8.

In the figure, ∠ADC = 130° and chord BC = chord BE, Find ∠CBE.

Solution :

ABCD is a cyclic quadrilateral.

We know that sum of opposite (RBSESolutions.com) angles of cyclic quadrilateral is 180°

∴ ∠ADC + ∠ABC = 180°

⇒ ∠ABC = 180° – ∠ADC

⇒ ∠ABC = 180° – 130°

⇒ ∠ABC = 50°

⇒ ∠OBC = 50°

In ΔBCO and ΔBEO

BC = BE (given)

∠BCO = ∠BEO [Angles opposite to equal sides]

BO = BO (common)

∴ By SAS congruence

ΔBCO = ΔBEO

∴ ∠OBC = ∠OBE

∴ ∠OBE = 50° [∠OBC = 50°]

Now ∠CBE = ∠CBO + ∠OBE

= 50° + 50°

= 100°

Thus ∠CBE = 100°

Question 9.

In fig. ∠ACB = 40°, find ∠OAB

Solution :

∴ ∠ACB = 40°

We know that angle subtended by arc of a (RBSESolutions.com) circle at center is double the angle at remaining part.

∴ ∠AOB = 2∠ACB

= 2 × 40°

= 80°

⇒ ∠AOB = 80°

∵ OA = OB = radius of circle

∴ In ∆AOB

∠OAB + ∠OBA + ∠AOB = 180°

∵ Angles opposite to equal sides are equal.

∴ ∠OAB = ∠OBA

⇒ ∠OAB + ∠OAB + 80° = 180°

⇒ 2∠OAB = 180° – 80°

⇒ ∠OAB = \(\frac { { 100 }^{ \circ } }{ 2 } \)

⇒ ∠OAB = 50°

Question 10.

In figure, AOB is diameter of circle and C,D and E are (RBSESolutions.com) any three points on semicircle find ∠ACD + ∠BED

Solution :

Joing AE and angle in semicircle ACHE is a cyclic quadrilateral opposite angles of cyclic quadrilateral are supplementary

∴ ∠AEB = 90° …(i)

∴ ∠ACD + ∠DEA = 180° …(ii)

Adding equation (i) and (ii)

∠AEB + ∠DEA + ∠ACD = 180° + 90°

∠BED + ∠ACD = 270°

[∵ ∠AEB + ∠DEA = ∠BED]

We hope the given RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 12 Circle Exercise 12.3, drop a comment below and we will get back to you at the earliest.

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