Circles

Exercise 17(A)

Question 1

In the given figure, O is the center of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC. Show your steps of working.

Answer

Join AC.

As, OA = OC = radius of circle.

Let ∠OAC = ∠OCA = x [Angles opposite to equal sides are equal]

We know that,

Sum of angles of triangle = 180°

∴ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + ∠AOC = 180°

⇒ ∠AOC = 180° - 2x

From figure,

⇒ ∠BAC = ∠BAO + OAC = 30° + x

⇒ ∠BCA = ∠BCO + OCA = 40° + x

Now, in ∆ABC

⇒ ∠ABC = 180° - ∠BAC - ∠BCA [Angle sum property of a triangle]

= 180° - (30° + x) - (40° + x)

= 180° - 30° - x - 40° - x

= 180° - 70° - 2x

= 110° - 2x

We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠AOC = 2∠ABC

⇒ 180° - 2x = 2(110° - 2x)

⇒ 180° - 2x = 220° - 4x

⇒ -2x + 4x = 220° - 180°

⇒ 2x = 40°

⇒ x = 20°.

Thus, ∠AOC = 180° - 2x = 180° - 2(20°)

= 180° - 40° = 140°.

Hence, ∠AOC = 140°.

Question 3

Given O is the centre of the circle and ∠AOB = 70°. Calculate the value of :

(i) ∠OCA,

(ii) ∠OAC.

Answer

We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠AOB = 2∠ACB

⇒ 70° = 2∠ACB

⇒ ∠ACB = $\dfrac{70}{2}$ = 35°.

(i) From figure,

∠OCA = ∠ACB = 35°.

Hence, ∠OCA = 35°.

(ii) Since, BC is a straight line,

∴ ∠AOB + ∠AOC = 180°

⇒ 70° + ∠AOC = 180°

⇒ ∠AOC = 180° - 70° = 110°.

In △OAC,

⇒ ∠OCA + ∠OAC + ∠AOC = 180°

⇒ ∠OAC + 35° + 110° = 180°

⇒ ∠OAC + 145° = 180°

⇒ ∠OAC = 180° - 145° = 35°.

Hence, ∠OAC = 35°.

Question 6

In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O₁ and O₂ are the centers of two circles.

In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O<sub>1</sub> and O<sub>2</sub> are the centers of two circles. Circles, Concise Mathematics Solutions ICSE Class 10.

Answer

We know that,

Angle in a semi-circle is a right angle.

∴ ∠DBA = ∠CBA = 90°

Adding both we get,

⇒ ∠DBA + ∠CBA = 180°

Hence proved that, D, B and C form a straight line.

Question 7

In the figure, given below, find :

(i) ∠BCD,

(ii) ∠ADC,

(iii) ∠ABC.

Show steps of your working.

Answer

(i) We know that,

Sum of opposite angles in a cyclic quadrilateral is 180°.

⇒ ∠BCD + ∠DAB = 180°

⇒ ∠BCD + 105° = 180°

⇒ ∠BCD = 180° - 105° = 75°.

Hence, ∠BCD = 75°.

(ii) Sum of co-interior angles in a trapezium = 180°.

⇒ ∠ADC + ∠DAB = 180°

⇒ ∠ADC + 105° = 180°

⇒ ∠ADC = 180° - 105° = 75°.

Hence, ∠ADC = 75°.

(iii) We know that, the sum of angles in a quadrilateral is 360°

So,

⇒ ∠ADC + ∠DAB + ∠BCD + ∠ABC = 360°

⇒ 75° + 105° + 75° + ∠ABC = 360°

⇒ ∠ABC = 360° - 255°

⇒ ∠ABC = 105°.

Hence, ∠ABC = 105°.

Question 8

In the given figure, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°; find :

(i) ∠ACB,

(ii) ∠OBC,

(iii) ∠OAB,

(iv) ∠CBA.

In the given figure, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°; find : Circles, Concise Mathematics Solutions ICSE Class 10.

Answer

Given, ∠AOB = 140° and ∠OAC = 50°

(i) We know that,

Angle at the center is double the angle at the circumference subtended by the same chord.

∠ACB = $\dfrac{1}{2}$ Reflex ∠AOB

= $\dfrac{1}{2}$ (360° - 140°)

= $\dfrac{1}{2} \times 220°$ = 110°.

Hence, ∠ACB = 110°.

(ii) We know that,

The sum of angles in a quadrilateral is 360°

In quadrilateral OBCA,

∠OBC + ∠ACB + ∠OAC + ∠AOB = 360°

⇒ ∠OBC + 110° + 50° + 140° = 360°

⇒ ∠OBC + 300° = 360°

⇒ ∠OBC = 360° - 300° = 60°.

Hence, ∠OBC = 60°.

(iii) Join AB.

In ∆AOB, we have

OA = OB (radius of circle)

So, ∠OBA = ∠OAB (As angles opposite to equal sides are equal)

By angle sum property of a triangle we get,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 2∠OAB + 140° = 180°

⇒ 2∠OAB = 40°

⇒ ∠OAB = $\dfrac{40}{2}$ = 20°

Hence, ∠OAB = 20°.

(iv) We already found, ∠OBC = 60°.

⇒ ∠OBC = ∠CBA + ∠OBA

⇒ 60° = ∠CBA + 20°

⇒ ∠CBA = 60° - 20° = 40°

Hence, ∠CBA = 40°.

Question 9

Calculate:

(i) ∠CDB,

(ii) ∠ABC,

(iii) ∠ACB.

Answer

(i) We know that,

Angles subtended by the same chord on the circle are equal.

∴ ∠CDB = ∠BAC = 49°.

Hence, ∠CDB = 49°.

(ii) We know that,

Angles subtended by the same chord on the circle are equal.

∠ABC = ∠ADC = 43°.

Hence, ∠ABC = 43°.

(iii) In △ABC,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ 43° + ∠ACB + 49° = 180°

⇒ ∠ACB + 92° = 180°

⇒ ∠ACB = 180° - 92° = 88°.

Hence, ∠ACB = 88°.

Question 12

Given: ∠CAB = 75° and ∠CBA = 50°. Find the value of ∠DAB + ∠ABD.

Answer

In ∆ABC, by angle sum property we have

⇒ ∠ACB + ∠CBA + ∠CAB = 180°

⇒ ∠ACB + 50° + 75° = 180°

⇒ ∠ACB + 125° = 180°

⇒ ∠ACB = 180° - 125° = 55°.

We know that,

Angles subtended by the same chord on the circle are equal.

⇒ ∠ADB = ∠ACB = 55°.

Now, taking ∆ABD

⇒ ∠DAB + ∠ABD + ∠ADB = 180° [Angle sum property]

⇒ ∠DAB + ∠ABD + 55° = 180°

⇒ ∠DAB + ∠ABD = 180° - 55°

⇒ ∠DAB + ∠ABD = 125°

Hence, ∠DAB + ∠ABD = 125°.

Question 13

ABCD is a cyclic quadrilateral in a circle with centre O. If ∠ADC = 130°, find ∠BAC.

Answer

We know that,

Angle in a semi-circle is 90°.

∠ACB = 90°.

We know that,

Sum of opposite angles of a cyclic quadrilateral = 180°.

⇒ ∠ABC = 180° - ∠ADC = 180° - 130° = 50°.

In △ACB,

⇒ ∠ACB + ∠CBA + ∠BAC = 180° [Angle sum property]

⇒ 90° + 50° + ∠BAC = 180°

⇒ ∠BAC + 140° = 180°

⇒ ∠BAC = 180° - 140° = 40°.

Hence, ∠BAC = 40°.

Question 15

In the following figure, O is the centre of the circle, ∠AOB = 60° and ∠BDC = 100°. Find ∠OBC.

Answer

We know that,

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

∠ACB = $\dfrac{1}{2}$ ∠AOB = $\dfrac{1}{2} \times 60°$ = 30°.

In △BDC,

⇒ ∠BDC + ∠DCB + ∠CBD = 180° [Angle sum property]

⇒ 100° + 30° + ∠CBD = 180° [From figure, ∠DCB = ∠ACB]

⇒ ∠CBD + 130° = 180°

⇒ ∠CBD = 180° - 130° = 50°.

From figure,

⇒ ∠OBC = ∠CBD = 50°.

Hence, ∠OBC = 50°.

Question 16

In cyclic quadrilateral ABCD, ∠DAC = 27°; ∠DBA = 50° and ∠ADB = 33°. Calculate :

(i) ∠DBC,

(ii) ∠DCB,

(iii) ∠CAB.

Answer

(i) We know that,

Angles in same segment are equal.

∠DBC = ∠DAC = 27°.

Hence, ∠DBC = 27°.

(ii) We know that,

Angles in same segment are equal.

∠ACB = ∠ADB = 33°.

and,

∠ACD = ∠ABD = 50°.

From figure,

⇒ ∠DCB = ∠ACD + ∠ACB = 50° + 33° = 83°.

Hence, ∠DCB = 83°.

(iii) In quad. ABCD,

⇒ ∠DAB + ∠DCB = 180° [As sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ ∠DAC + ∠CAB + ∠DCB = 180°

⇒ 27° + 83° + ∠CAB = 180°

⇒ ∠CAB + 110° = 180°

⇒ ∠CAB = 180° - 110° = 70°.

Hence, ∠CAB = 70°.