Kolkata board math's chapter 6 solution

Theorems Related To Angles In A Circle

Class 10^th Mathematics West Bengal Board Solution

Let Us Work Out 7.1

1. O is the circumcentre of the isosceles triangle ABC, whose AB = AC, the point A and BC…
2. In the adjoining figure, if O is the centre of circumcircle of ΔABC. And ∠AOC = 110°,…
3. ABCD is a cyclic quadrilateral of a circle with centre O; DC is extended to the point…
4. In the adjacent figure O is the centre of the circle; ∠AOD = 40° and ∠ACB = 35°; let us…
5. O is the centre of circle in the picture beside, if ∠APB = 80°, let us find the sum of…
6. Like the adjoining figure, we draw two circles with centres C and D which intersect…
7. If the circumcentre of triangle ABC is O; let us prove that ∠OBC + ∠BAC = 90°.…
8. Each of two equal circles passes through the centre of the other and the two circles…
9. S is the centre of the circumcircle of ΔABC and if AD ⊥ BC, let us prove that ∠BAD =…
10. Two chords AB and CD of a circle with centre O intersect each other at the points P,…
11. If two chords AB and CD of a circle with centre O, when produced intersect each other…
12. We drew a circle with the point A of quadrilateral ABCD as centre which passes through…
13. O is the circumcentre of Δ ABC and OD is perpendicular on the side BC; let us prove…
14. Q14A1 In the adjoining figure, if ‘O’ is the centre of circle and PQ is a diameter, then…
15. Q14A2 In the adjoining figure, if O is the centre of circle, the then the value of x is A.…
16. Q14A3 In the adjoining figure, if O is centre of circle and BC is the diameter then the…
17. Q14A4 O is the circumcentre of Δ ABC and ∠OAB = 50°, then the value of ∠ACB isA. 50 B. 100…
18. Q14A5 In the adjoining figure, if O is centre of circle, the value of ∠POR is q A. 20 B.…
19. Let us write whether the following statements are true of false” (i) In the adjoining…
20. Let us fill in the blanks (i) The angle at the centre is ______ the angle on the…
21. In the adjoining figure, O is the centre of the circle, if ∠ACB = 30°, ∠ABC = 60°,…
22. O is circumcentre of the triangle ABC and D is the mid point of the side BC. If ∠BAC…
23. Three points A, B and C lie on the circle with centre O in such a way that AOCB is a…
24. O is the circumcentre of isosceles triangle ABC and ∠ABC = 120°; if the length of the…
25. Two circles with centres A and B interest each other at the points C and D. The…

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Let Us Work Out 7.1

Question 1.

O is the circumcentre of the isosceles triangle ABC, whose AB = AC, the point A and BC are on opposite us write by calculating, the values of ∠ABC and ∠ABO.

Answer:

Since it is given that ∠BOC = 100° and by the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

In ΔABC, as AB = BC, and we know that angle opposite to equal sides are equal.

⇒ ∠ACB = ∠ABC = t

As the sum of all angles in a triangle is equal to 180°

⇒ 2t + ∠BAC = 180

⇒ 2t = 180 – 50

⇒ t = 65°

⇒ ∠ABC = 65°

Similarly, in ΔBOC, BO = OC, applying the same principle as above, we get,

⇒ 2∠OBC + ∠BOC = 180

⇒ 2∠OBC = 180 – 100

⇒ ∠OBC = 40°

Also, ∠ABO = ∠ABC - ∠OBC

⇒ ∠ABO = 65 – 40 = 15°

The values of ABC and ABO are 65° and 15°.

Question 2.

In the adjoining figure, if O is the centre of circumcircle of ΔABC. And ∠AOC = 110°, let us write by calculating, the value of ∠ABC.

Answer:

Since it is given that ∠AOC = 110° and by the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

Since APCB forms a cyclic quadrilateral and we know that sum of opposite sides of a cyclic quadrilateral is equal to 180°

⇒ ∠APC + ∠ABC = 180

⇒ ∠ABC = 180 – 55 = 125°

Question 3.

ABCD is a cyclic quadrilateral of a circle with centre O; DC is extended to the point P. If ∠BCP = 180°, let us write by calculating, the value of ∠BOD.

Answer:

Since ∠BCP = 108° and DCP is a straight line,

⇒ ∠BCD = 180 – 108

∠BCD = 72°

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

∴ value of ∠BOD is 144°.

Question 4.

In the adjacent figure O is the centre of the circle; ∠AOD = 40° and ∠ACB = 35°; let us write by calculating the value of ∠BCO and ∠BOD, and answer with reason.

Answer:

As ∠ACB = 35° and by the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

As ∠BOD = ∠AOD + ∠AOB

⇒ ∠BOD = 40 + 70 = 110°

Also since DOC forms a straight line,

⇒ ∠BOC = 180 - ∠BOD = 180 – 110 = 70°

In ΔBOC, OB = OC and as we know that angle opposite to equal sides are equal.

⇒ ∠OCB = ∠OBC

Sum of angles in a triangle is equal to 180.

⇒ 2∠BCO = 180 – 70

⇒ ∠BCO = 55°

Question 5.

O is the centre of circle in the picture beside, if ∠APB = 80°, let us find the sum of the measures of ∠AOB and ∠COD and answer with reason.

Answer:

In ΔAPB, sum of all angle is equal to 180°

⇒ ∠PAB + ∠ABP = 180 - ∠APB

⇒ ∠PAB + ∠ABP = 180 – 80 = 100 --------- (1)

Also by the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

and

Substituting above values in eq. (1)

⇒ ∠AOD + ∠BOC = 200

Also, as the sum of angles around the point is equal to 360°

⇒ ∠AOD + ∠BOC + ∠AOB + ∠COD = 360

⇒ ∠AOB + ∠COD = 360 – 200 = 160°

Question 6.

Like the adjoining figure, we draw two circles with centres C and D which intersect each other at the points A and B. We draw a straight line through the point A which intersects the circle with centre C at the point P and the circle with centre D at the point Q.

Let us prove that (i) ∠PBQ = ∠CAD (ii) ∠BPC = ∠BQD

Answer:

(i) In ΔACB and ΔADB,

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

and

--------- (1)

In ΔPCA, ∠PAC + ∠CPA = 180 - ∠PCA

⇒ 2∠CAP = 180 - ∠PCA [As ∠PAC = ∠CPA]

Similarly, in ΔADQ,

⇒ 2∠QAD = 180 - ∠ADQ [As ∠QAD = ∠AQD]

Since, PAQ is a straight line,

⇒ ∠CAD = 180 – (∠CAP + ∠QAD)

On substituting values in above equation, we get,

------------ (2)

From (1) and (2), we get, ∠CAD = ∠PBQ ---------- (3)

Hence, proved.

(ii) Also, as ∠CBA = ∠CAB and ∠ABD = ∠BAD

⇒ ∠CBA + ∠ABD = ∠CAB + ∠BAD

⇒ ∠CAD = ∠CBD ------------- (4)

From equations (3) and (4), we get,

∠PBQ = ∠CBD

⇒ ∠PBQ - ∠CBD = 0

⇒ ∠PBC - ∠QBD = 0

⇒ ∠PBC = ∠QBD ---------------- (5)

As PC = CB and we know that angles opposite to equal sides are equal

⇒ ∠BPC = ∠PBC ----------------- (6)

As BD = DQ and we know that angles opposite to equal sides are equal.

⇒ ∠BQD = ∠QBD ----------------- (7)

From equations (5), (6) and (7)

∠BPC = ∠BQD

Question 7.

If the circumcentre of triangle ABC is O; let us prove that ∠OBC + ∠BAC = 90°.

Answer:

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

As OB = OC and we know that angles opposite to equal sides are equal.

⇒ ∠OBC = ∠OCB

In ΔOBC, as sum of all sides of a triangle is equal to 180°.

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ 2∠OBC + ∠BOC = 180°

⇒ 2∠OBC + 2∠BAC = 180°

⇒ ∠OBC + ∠BAC = 90°

Hence, Proved.

Question 8.

Each of two equal circles passes through the centre of the other and the two circles intersect each other at the points A and B. If a straight line through the point A intersects the two circles at points C and D, let us prove that ΔABCD is an equilateral triangle.

Answer:

As the circles with centre X and Y are equal, the radius of both the circles are equal.

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

------------ (1)

Since AXBC is a quadrilateral triangle, therefore the sum of opposite sides of a quadrilateral is equal to 180°

⇒ ∠AYB + ∠ACB = 180 --------- (2)

Also, by using above theorem, we get

⇒ ∠AYB = 2∠ADB ------------ (3)

In ΔAXB and ΔAYB,

AX = AY (radius of equal circles)

BX = BY (radius of equal circles)

AB = AB (common)

⇒ ΔAXB ≅ ΔAYB, by SSS congruency

∴ ∠AXB = ∠AYB ------------ (4)

From equation (1), (2) and (4) we get,

⇒ 3∠ACB = 180

⇒ ∠ACB = 60°

From equation (1), (3) and (4) we get,

∠ACB = ∠ADB = 60°

In ΔBCD, sum of all angles of a triangle is equal to 180°

⇒ ∠ACB + ∠ADB + ∠CBD = 180°

⇒ ∠CBD = 60°

As each angle in ΔBCD is equal to 60°, therefore ΔBCD is an equilateral triangle.

Question 9.

S is the centre of the circumcircle of ΔABC and if AD ⊥ BC, let us prove that ∠BAD = ∠SAC.

Answer:

Since AD is perpendicular to BC, ∠ADB = 90°

⇒ ∠ABD + ∠BAD = 90 ----------- (1)

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

------------ (2)

From equation (1) and (2), we get,

------------- (3)

In ΔASC, as AS = SC and we know that angles opposite to equal sides are equal.

⇒ ∠SAC = ∠SCA

Also, ∠SAC + ∠SCA + ∠ASC = 180°

⇒ 2∠SAC + ∠ASC = 180°

⇒ ∠ASC = 180° - 2∠SAC ------------ (4)

From equation (3) and (4), we get,

⇒ ∠BAD = ∠SAC

Hence, proved.

Question 10.

Two chords AB and CD of a circle with centre O intersect each other at the points P, let us prove that ∠AOD + ∠BOC = 2∠BPC.

If AOD and BOC are supplementary to each other, let us prove that the two chords are perpendicular to each other.

Answer:

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

Similarly,

∠BOC = 2∠BAC

Adding above equations, we get,

⇒ ∠AOD+∠BOC = 2(∠BAC + ∠DCA) ----------- (1)

In ΔAPC, ∠PAC + ∠PCA = ∠BPC by exterior angles property.

⇒ ∠BAC + ∠DCA = ∠BPC

Hence, proved.

If AOD and BOC are supplementary to each other,

⇒ ∠BAC + ∠DCA = 90

And from above theorem, ∠BPC = 90°

Question 11.

If two chords AB and CD of a circle with centre O, when produced intersect each other at the point P, let us prove that ∠AOC - ∠BOD = 2 ∠BPC.

Answer:

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

Similarly, the angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

⇒ ∠BOD = 2∠BCD.

In ΔBPC, ∠ABC = ∠BPC + ∠BCP

On substituting the values of ∠ABC and ∠BCP in above

equation, we get,

⇒ ∠AOC - ∠BOD = 2 ∠BPC

Hence, proved.

Question 12.

We drew a circle with the point A of quadrilateral ABCD as centre which passes through the points B,C and D. Let us prove that ∠CBD + ∠ CDB =

Answer:

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

∠BAD = 2 ∠BED

Since BCDE forms cyclic quadrilateral, sum of opposite angles in a cyclic quadrilateral is equal to 180°

⇒ ∠BCD + ∠BED = 180

In ΔBCD, sum of all angles is equal to 180°

⇒ ∠BCD + ∠CBD + ∠CDB = 180

Hence, proved.

Question 13.

O is the circumcentre of Δ ABC and OD is perpendicular on the side BC; let us prove that ∠BOD = ∠BAC

Answer:

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

∠BOC = 2 ∠BAC ------------ (1)

In ΔBOD and ∠COD,

∠BDO = ∠CDO = 90° (given)

OD = OD (common)

OB = OC (radius)

Therefore, ΔBOD ≅ ∠COD by RHS congruency

⇒ ∠BOD = ∠COD ------------------- (2)

From (1) and (2), we get,

2 ∠BOD = 2 ∠BAC

⇒ ∠BOD = ∠BAC

Hence, proved.

Question 14.

In the adjoining figure, if ‘O’ is the centre of circle and PQ is a diameter, then the value of x is


A. 140

B. 40

C. 80

D. 20

Answer:

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

∠POR = 2 ∠PSR

⇒ ∠PSR = 70°

As PQ is diameter, angle subtended by the diagonal at any point on the circle is equal to 90°

⇒ x = 90 – 70 = 20°

Question 15.

In the adjoining figure, if O is the centre of circle, the then the value of x is

A. 70

B. 60

C. 40

D. 200

Answer:

∠QOR = 360 - 140 – 80

⇒ ∠QOR = 140°

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

∠QOR = 2x

⇒ x = 70°

Question 16.

In the adjoining figure, if O is centre of circle and BC is the diameter then the value of x is


A. 60

B. 50

C. 100

D. 80

Answer:

In ΔABO, 2∠BAO + ∠BOA = 180

⇒ ∠BOA = 80°

⇒ ∠AOC = 180 – 80 = 100

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

⇒ ∠AOC = 2x

⇒ x = 50°

Question 17.

O is the circumcentre of Δ ABC and ∠OAB = 50°, then the value of ∠ACB is
A. 50

B. 100

C. 40

D. 80

Answer:

In ΔABO, the value of ∠AOB = 80.

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

⇒ ∠AOB = 2 ∠ACB

⇒ ∠ACB = 40°

Question 18.

In the adjoining figure, if O is centre of circle, the value of ∠POR is


A. 20

B. 40

C. 60

D. 80

Answer:

In ΔORQ,

∠ROQ = 180 – 40 – 40

⇒ ∠ROQ = 100°

In ΔPOQ,

∠POQ = 180 – 10 – 10

⇒ ∠ROQ = 160°

∠POR = 160 – 100

⇒ ∠POR = 60°

Question 19.

Let us write whether the following statements are true of false”

(i) In the adjoining figure, if O is centre of circle, then ∠AOB = 2 ∠ACD



(ii) The point O lies within the triangular region ABC in such a way that OA = OB and ∠AOB = 2∠ACB. If we draw a circle with centre O and length of radius OA, then the point c lies on the circle.

Answer:

(i) False

For the given condition to happen, the angles should be subtended by the same arc only.

(ii) True

The given statement is true, due to the following theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle. Therefore, point C lies on circle.

Question 20.

Let us fill in the blanks

(i) The angle at the centre is ______ the angle on the circle, subtended by the same arc.

(ii) The length of two chord AB and CD are equal of a circle with centre O. If ∠APB and ∠DQC are angles on the circle, then the values of the two angles are_______.

(iii) If O is the circumcentre of equilateral triangle, then the value of the front angle formed by any side of the triangle is_______.

Answer:

(i) Double

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

(ii) Equal

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

(iii) 60°

The front angle is the angle subtended by the arc at any point on the circle in its opposite arc.

Question 21.

In the adjoining figure, O is the centre of the circle, if ∠ACB = 30°, ∠ABC = 60°, ∠DAB = 35° and ∠DBC = x°, the value of x is

Answer:

Let P be any Point in major arc of circle.

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

⇒ x = 2 ∠APC

AS APCB is a cyclic quadrilateral, so sum of opposite sides is equal to 180°

⇒ ∠APC + ∠ABC = 180

⇒ x = 120°

Also, ABCO is a quadrilateral whose sum of interior angles is equal to 360°.

⇒ y = 360 – x – 120 – 30

⇒ y = 90°

Question 22.

O is circumcentre of the triangle ABC and D is the mid point of the side BC. If ∠BAC = 40°, let us find the value of ∠BOD.

Answer:

Since D is the midpoint of BC,

In ΔBOD and ΔCOD, we have,

BO = CO (radius)

OD = OD (common)

BD = DC (given D as midpoint)

∴ ΔBOD ≅ ΔCOD by SSS congruency

⇒ ∠BOD = ∠COD ------------- (1)

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

∠BOC = 2 ∠BAC

⇒ ∠BOD + ∠COD = 2 ∠BAC

From equation (1), we get,

⇒ 2 ∠BOD = 2 ∠BAC

⇒ ∠BOD = ∠BAC = 40°

Question 23.

Three points A, B and C lie on the circle with centre O in such a way that AOCB is a parallelogram, let us calculate the value of ∠AOC.

Answer:

As OABC is a parallelogram, therefore, opposite angles will be equal.

⇒ ∠AOC = ∠ABC --------- (1)

Also, ABCD is a cyclic quadrilateral, therefore sum of opposite angles must be 180°

⇒ ∠ADC + ∠ABC = 180 ---------- (2)

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

∠AOC = 2 ∠ADC -------- (3)

From (1), (2) and (3), we get,

⇒ ∠AOC = 120°

Question 24.

O is the circumcentre of isosceles triangle ABC and ∠ABC = 120°; if the length of the radius of the circle is 5 cm, let us find the value of the side AB.

Answer:

Since ∠ABC = 120° and therefore the angle subtended by arc ABC in major arc will be:

⇒ ∠APB = 180 – 120( By using Property of cyclic quadrilateral, where P is any point on circle)

⇒ ∠APB = 60°

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

∠AOC = 2 ∠APB = 120°

In ΔAOB and ΔCOB,

AO = CO (radius)

OB = OB (given)

AB = BC (given)

∴ ΔAOB ≅ ΔCOB by SSS congruency

⇒ ∠ABO = ∠CBO = 60°

Also, ∠AOB = ∠COB = 60° .

This shows ΔOAB is an equilateral triangle.

⇒ AB = OA = 5 cm

Question 25.

Two circles with centres A and B interest each other at the points C and D. The centre lies on the circle with centre A. If ∠CQD = 70°, let us find the value of ∠CPD.

Answer:

In circle with centre B, by the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

∠CBD = 2 ∠CQD = 140°

Now, since B lies on the circle with centre A, therefore, quadrilateral CPDB is cyclic quadrilateral.

We also know that, the sum of opposite sides in a cyclic quadrilateral is equal to 180°

⇒ ∠CPD + ∠CBD = 180°

On substituting the value in above equation, we get,

⇒ ∠CPD = 180 – 140

⇒ ∠CPD = 40°