# ICSE Solutions for Chapter 17 Circles Class 10 Mathematics

**Question 1: In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30˚ and 40˚ respectively. Find angle AOC show your steps of working.**

**Solution 1:**

Let ∠OAC = ∠OCA = x (say)

∴ ∠AOC = 180˚ – 2x

Also, ∠BAC = 30˚ + x

∠BCA = 40˚ + x

In Î”ABC,

∠ABC = 180˚ - ∠BAC - ∠BCA

= 180 – (30˚ + x) – (40˚ + x) = 110˚ – 2x

Now, ∠AOC = 2∠ABC

**...(Angle at the centre is double the angle at the circumference subtended by the same chord)**

⇒ 180˚ – 2x = 2(110˚ – 2x)

⇒ 2x = 40˚

∴ x = 20˚

∴ ∠AOC = 180˚ – 2×20˚ = 140˚

**Question 2: In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°**

**(i) Prove that AC is a diameter of the circle.**

**(ii) Find ∠ACB.**

**Solution 2:**(i) In Î”ABD,

∠DAB + ∠ABD + ∠ADB = 180°

⇒ 65° +70° + ∠ADB = 180°

⇒ 135° + ∠ADB = 180°

⇒ ADB = 180° - 135° = 45°

Now, ∠ADC = ∠ADB + ∠BDC = 45° + 45°= 90°

Since, ∠ADC is the angle of semicircle, so AC is a diameter of the circle.

(ii) ∠ACB = ∠ADB

⇒ ∠ACB = 45°

Here, ∠AOB = 2∠ACB

⇒ ∠ACB = 70°/2 = 35°

Now, OC = OA

⇒ ∠OCA = ∠OAC = 35˚

Here, b = ½×130˚

⇒ b = 65˚

Now, a + b = 180°

⇒ a = 180° - 65° = 115˚

(ii) Here, c = ½ Reflex (112˚)

⇒ c = ½×(360° - 112˚) = 124˚

∴ ∠ACB = 1/2 ∠AOB

Since AB is the side of a regular hexagon,

∠AOB = 60°

**….(angles in the same segment of the circle)**⇒ ∠ACB = 45°

**Question 3: Given O is the centre of the circle and ∠AOB = 70˚. Calculate the value of:****(i) ∠OCA; (ii) ∠OAC.**

**Solution 3:****...(Angle at the centre is double the angle at the circumference subtended by the same chord)**⇒ ∠ACB = 70°/2 = 35°

Now, OC = OA

**...(Radii of same circle)**⇒ ∠OCA = ∠OAC = 35˚

**Question 4: In each of the following figures, O is the centre of the circle. Find the values of a, b and c.**

**Solution 4:****...(Angle at the centre is double the angle at the circumference subtended by the same chord)**⇒ b = 65˚

Now, a + b = 180°

**...(Opposite angles of a cyclic quadrilateral are supplementary)**⇒ a = 180° - 65° = 115˚

(ii) Here, c = ½ Reflex (112˚)

**...(Angle at the centre is double the angle at the circumference subtended by the same chord)**⇒ c = ½×(360° - 112˚) = 124˚

**Question 5: In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of:****(i) ∠AOB,****(ii) ∠ACB,****(iii) ∠ABC.**

**Solution 5:**(i) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.∴ ∠ACB = 1/2 ∠AOB

Since AB is the side of a regular hexagon,

∠AOB = 60°

(ii) ∠AOB = 60˚

⇒ ∠ACB = ½×60° = 30°

(iii) Since AC is the side of a regular octagon,

∠AOC = 360˚/8 = 45°

Again, Arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.

⇒ ∠ABC = 1/2∠AOC

⇒ ABC = 45°/2 = 22.5˚

Arc AE subtends ∠AOE at the centre and ∠ADE at the remaining part of the circle.

∴ ∠ADE = ½ ∠AOE

= ½×72˚

= 36˚

∠ADC = ∠ADB + ∠BDC

= 36° + 36° + 72˚

∴ ∠ADE : ∠ADC = 36°:72° = 1:2

(i) ∠BCD + ∠BAD = 180°

(Sum of opposite angles of a cyclic quadrilateral is 180°)

⇒ ∠BCD = 180° - 105° = 75˚

∠AOC = 360˚/8 = 45°

Again, Arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.

⇒ ∠ABC = 1/2∠AOC

⇒ ABC = 45°/2 = 22.5˚

**Question 6: In a regular pentagon ABCDE, inscribed in a circle; find ratio between angle EDA and angel ADC.**

**Solution 6:**∴ ∠ADE = ½ ∠AOE

= ½×72˚

= 36˚

**[central angle of a regular pentagon at O]**∠ADC = ∠ADB + ∠BDC

= 36° + 36° + 72˚

∴ ∠ADE : ∠ADC = 36°:72° = 1:2

**Question 7: In the figure, given below, find****(i) ∠BCD, (ii) ∠ADC, (iii) ∠ABC. Show steps of your working.**

**Solution 7:**(Sum of opposite angles of a cyclic quadrilateral is 180°)

⇒ ∠BCD = 180° - 105° = 75˚

(ii) Now, AB || CD

∴ ∠BAD + ∠ADC = 180°

(Interior angles on same side of parallel lines is 180°)

⇒ ∠ADC = 180° - 105° = 75˚

∴ ∠BAD + ∠ADC = 180°

(Interior angles on same side of parallel lines is 180°)

⇒ ∠ADC = 180° - 105° = 75˚

(iii) ∠ADC + ∠ABC = 180°

(Sum of opposite angles of a cyclic quadrilateral is 180°)

⇒ ∠ABC = 180° - 75° = 105˚

Here, ∠ACB = ½ Reflex (∠AOB) = ½(360° - 140°) = 110°

(Angle at the centre is double the angle at the circumference subtended by the same chord)

Now, OA = OB

∴ ∠OBA = ∠OAB = (180° - 140˚)/2 = 20˚

∴ ∠CAB = 50° - 20° = 30˚

In Î”CAB,

∠CBA = 180° - 110° - 30° = 40°

∴ ∠OBC = ∠CBA + ∠OBA = 40° + 20° = 60˚

Here, ∠CDB = ∠BAC = 49˚

∠ABC - ∠ADC = 43°

By angle-sum property of a triangle,

∠ACB = 180° - 49° - 43° = 88°

(i) By angle – sum property of triangle ABD,

∠ADB = 180° - 75˚ - 58˚ = 47˚

∠BDC = ∠ADC - ∠ADB = 770 - 47° = 30°

(Sum of opposite angles of a cyclic quadrilateral is 180°)

⇒ ∠ABC = 180° - 75° = 105˚

**Question 8: In the figure, given below, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°; find :****(i) ∠ACB; (ii) ∠OBC; (iii) ∠OAB ; (iv) ∠CBA.****Solution 8:**(Angle at the centre is double the angle at the circumference subtended by the same chord)

Now, OA = OB

**...(Radii of same circle)**∴ ∠OBA = ∠OAB = (180° - 140˚)/2 = 20˚

∴ ∠CAB = 50° - 20° = 30˚

In Î”CAB,

∠CBA = 180° - 110° - 30° = 40°

∴ ∠OBC = ∠CBA + ∠OBA = 40° + 20° = 60˚

**Question 9: Calculate :****(i) ∠CDB, (ii) ∠ABC, (iii) ∠ACB.**

**Solution 9:**∠ABC - ∠ADC = 43°

**...(Angles subtended by the same chord on the circle are equal)**By angle-sum property of a triangle,

∠ACB = 180° - 49° - 43° = 88°

**Question 10: In the figure given below, ABCD is a cyclic quadrilateral in which ∠BAD= 75°: ∠ABD = 58° and ∠ADC = 77˚.****Find: (i) ∠BDC, (ii) ∠BCD, (iii) ∠BCA.**

**Solution 10:**∠ADB = 180° - 75˚ - 58˚ = 47˚

∠BDC = ∠ADC - ∠ADB = 770 - 47° = 30°

(ii) ∠BAD + ∠BCD = 180°

⇒ ∠BCD = 180° - 75˚ = 105˚

**(Sum of opposite angles of a cyclic quadrilateral is 180°)**⇒ ∠BCD = 180° - 75˚ = 105˚

(iii) ∠BCA = ∠ADB = 47˚

(Angles subtended by the same chord on the circle are equal)

(Angles subtended by the same chord on the circle are equal)