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Theorems Related To Angles In A Circle

Class 10th 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…
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°