Circles — Exercise 17(A)

In the given figure, O is center of the circle and ∠B = 55°. The angle A is equal to :

1. 55°

2. 35°

3. 45°

4. 50°

Answer

From figure,

In △ OCB,

OB = OC (Radius of common circle)

We know that,

Angles opposite to equal sides of a triangle are equal.

∴ ∠C = ∠B = 55°

By angle sum property of triangle,

⇒ ∠C + ∠B + ∠O = 180°

⇒ 55° + 55° + ∠O = 180°

⇒ ∠O + 110° = 180°

⇒ ∠O = 180° - 110° = 70°.

We know that,

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

⇒ ∠O = 2∠A

⇒ ∠A = $\dfrac{∠O}{2} = \dfrac{70°}{2}$ = 35°.

Hence, Option 2 is the correct option.

In the given figure, O is center of the circle and angle OBA = 50°. The angle P is :

1. 50°

2. 80°

3. 40°

4. 60°

Answer

From figure,

In △ OAB,

OA = OB (Radius of common circle)

We know that,

Angles opposite to equal sides of a triangle are equal.

∴ ∠A = ∠B = 50°

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠O = 180°

⇒ 50° + 50° + ∠O = 180°

⇒ ∠O + 100° = 180°

⇒ ∠O = 180° - 100° = 80°.

We know that,

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

⇒ ∠O = 2∠P

⇒ ∠P = $\dfrac{∠O}{2} = \dfrac{80°}{2}$ = 40°.

Hence, Option 3 is the correct option.

In the given figure, chord AB = chord PB and angle C = 50°. The angle PAB is equal to :

1. 65°

2. 50°

3. 75°

4. 60°

Answer

We know that,

Angle in same segment are equal.

∴ ∠APB = ∠ACB = 50°

In △ PAB,

PB = AB (Given)

We know that,

Angles opposite to equal sides of a triangle are equal.

∴ ∠PAB = ∠APB = 50°.

Hence, Option 2 is the correct option.

O' and O" are centers of two circles which intersect each other at points A and B. Then :

1. BC = BD

2. BC is larger than BD.

3. BC is smaller than BD.

4. C, B and D are collinear.

Answer

We know that,

Angle in semicircle is a right angle.

From figure,

AC and AD are diameter.

∴ ∠ABC = ∠ABD = 90°.

From figure,

⇒ ∠ABC + ∠ABD = 90° + 90° = 180°.

∴ CBD is a straight line, or points C, B and D are collinear.

Hence, Option 4 is the correct option.

In the given figure, O is center of the circle, AB || DC and ∠ACD = 32°, ∠DAB is equal to :

1. 122°

2. 148°

3. 90°

4. none of the above

Answer

Join DA.

From figure,

In △DAC,

∠DAC = 90° (Angle in semicircle is a right angle)

As, alternate angles are equal.

∴ ∠CAB = ∠ACD = 32°

From figure,

∠DAB = ∠DAC + ∠CAB = 90° + 32° = 122°.

Hence, Option 1 is the correct option.

In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°

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

(ii) Find ∠ACB

Answer

(i) In ∆ABD,

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

⇒ 65° + 70° + ∠ADB = 180°

⇒ 135° + ∠ADB = 180°

⇒ ∠ADB = 180° - 135° = 45°

From figure,

⇒ ∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°.

As the angle in a semicircle is a right angle,

∴ Arc ADC is semi-circle and AC is the diameter.

Hence, proved that AC is the diameter.

(ii) We know that,

Angles in the same segment of a circle are equal.

∴ ∠ACB = ∠ADB = 45°.

Hence, ∠ACB = 45°.

In the following figure, O is the center of the circle. Find the value of a.

Answer

We know that,

Angle in semi-circle is a right angle.

∴ ∠BAD = 90°

From figure,

⇒ ∠BAD + ∠ADB + ∠DBA = 180°

⇒ 90° + ∠ADB + 35° = 180°

⇒ ∠ADB + 125° = 180°

⇒ ∠ADB = 180° - 125° = 55°.

We know that,

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

∴ a = ∠ADB = 55°.

Hence, a = 55°.

In the following figure, O is the center of the circle. Find the value of b.

Answer

Mark the point of intersection of AC and BD as E:

Since, AC is a straight line.

∴ ∠AEB + ∠BEC = 180°

⇒ 120° + ∠BEC = 180°

⇒ ∠BEC = 180° - 120° = 60°.

In △BEC,

⇒ ∠BEC + ∠ECB + ∠CBE = 180°

⇒ 60° + ∠ECB + 25° = 180°

⇒ ∠ECB + 85° = 180°

⇒ ∠ECB = 180° - 85° = 95°.

We know that,

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

∴ b = ∠ACB

From figure,

∠ACB = ∠ECB = 95°.

Hence, b = 95°.

In the following figure, O is the center of the circle. Find the value of c.

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 = 2 x 50° = 100°.

Also, OA = OB = radius of circle.

∴ ∠OBA = ∠OAB = c [As angles opposite to equal sides are equal]

In △OAB,

⇒ ∠OAB + ∠AOB + ∠OBA = 180°

⇒ 2∠OAB + 100° = 180° [∵ ∠OBA = ∠OAB and ∠AOB = 100°]

⇒ 2c + 100° = 180°

⇒ 2c = 180° - 100°

⇒ 2c = 80°

⇒ c = $\dfrac{80}{2}$ = 40°.

Hence, c = 40°.

In the following figure, O is the center of the circle. Find the value of d.

Answer

We know that,

Angle in semi-circle is a right angle.

∴ ∠APB = 90°

In △APB,

⇒ ∠APB + ∠PBA + ∠PAB = 180°

⇒ 90° + 45° + ∠PAB = 180°

⇒ ∠PAB = 180° - 135° = 45°.

We know that,

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

∴ d = ∠PAB = 45°.

Hence, d = 45°.

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°.

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°.

In the figure given alongside, AOB is a diameter of the circle and ∠AOC = 110°, find ∠BDC.

Answer

Join AD.

We know that,

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

∠ADC = $\dfrac{1}{2}$∠AOC = $\dfrac{1}{2}$ x 110° = 55°.

Also, we know that

Angle in the semi-circle is a right angle.

∠ADB = 90°

From figure,

∠BDC = ∠BDA - ∠ADC = 90° - 55° = 35°.

Hence, ∠BDC = 35°.

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°.

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°.

In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°, find the number of degrees in :

(i) ∠DCE,

(ii) ∠ABC.

Answer

(i) From figure,

∠CED = 90° [As angle in semi-circle = 90°]

In △CED,

⇒ ∠CED + ∠CDE + ∠DCE = 180° [By angle sum property of triangle]

⇒ 90° + 40° + ∠DCE = 180°

⇒ ∠DCE + 130° = 180°

⇒ ∠DCE = 180° - 130° = 50°.

Hence, ∠DCE = 50°.

(ii) We know that,

An exterior angle is equal to sum of two opposite interior angles.

In ∆BOC,

⇒ ∠AOC = ∠OCB + ∠OBC

⇒ ∠OBC = ∠AOC - ∠OCB

⇒ ∠OBC = ∠AOC - ∠DCE

⇒ ∠OBC = 80° - 50° = 30°.

From figure,

∠ABC = ∠OBC = 30°.

Hence, ∠ABC = 30°.

In the figure given alongside, AB || CD and O is the center of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.

Answer

Join AC, BD and CB.

As angle is a semicircle is a right angle.

∴ ∠CAD = 90° and ∠CBD = 90°

Given, AB || CD

So, ∠BAD = ∠ADC = 25° [Alternate angles are equal]

From figure,

∠BAC = ∠BAD + ∠CAD = 25° + 90° = 115°.

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

In quadrilateral ACDB,

⇒ ∠CDB + ∠BAC = 180°

⇒ ∠CDA + ∠ADB + ∠BAC = 180°

⇒ 25° + ∠ADB + 115° = 180°

⇒ ∠ADB = 180° - 115° - 25° = 40°.

As angles in same segment are equal.

∴ ∠AEB = ∠ADB = 40°.

Hence, ∠AEB = 40°.

AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find :

(i) ∠PRB,

(ii) ∠PBR,

(iii) ∠BPR.

Answer

(i) We know that,

Angles in same segment are equal.

∠PRB = ∠PAB = 35°.

Hence, ∠PRB = 35°.

(ii) From figure,

∠BPA = 90° [Angle in semi-circle is a right angle.]

⇒ ∠BPA + ∠BPQ = 180° [Linear pairs]

⇒ 90° + ∠BPQ = 180°

⇒ ∠BPQ = 180° - 90° = 90°.

Exterior angle of a triangle is equal to the sum of two opposite interior angles.

⇒ ∠PBR = ∠BPQ + ∠BQP = 90° + 25° = 115°.

Hence, ∠PBR = 115°.

(iii) In △ABP,

⇒ ∠BPA + ∠PAB + ∠ABP = 180°

⇒ 90° + 35° + ∠ABP = 180°

⇒ 125° + ∠ABP = 180°

⇒ ∠ABP = 180° - 125° = 55°.

From figure,

∠ABR = ∠PBR - ∠ABP = 115° - 55° = 60°.

∠APR = ∠ABR = 60°. [Angles in same segment are equal]

In △BPR,

⇒ ∠BPR + ∠PRB + ∠PBR = 180°

⇒ ∠BPR + 35° + 115° = 180°

⇒ 150° + ∠BPR = 180°

⇒ ∠BPR = 180° - 150° = 30°.

Hence, ∠BPR = 30°.

In the given figure, A is the center of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that : ∠BCD = 2∠ABE.

Answer

∠BAD = 2∠BED [Angle at the center is double the angle at the circumference subtended by the same chord.]

Since, CDE is a straight line and CD || AB.

∴ AB || ED.

⇒ ∠BED = ∠ABE [Alternate angles are equal]

Multiplying above equation by 2 we get,

⇒ 2∠BED = 2∠ABE

⇒ ∠BAD = 2∠ABE ...............(1)

ABCD is a parallelogram.

⇒ ∠BAD = ∠BCD [Opposite angles of a paralellogram are equal] ..........(2)

From (1) and (2) we get,

⇒ ∠BCD = 2∠ABE.

Hence, proved that ∠BCD = 2∠ABE.

In the given figure, I is the incenter of △ABC, BI when produced meets the circumcircle of △ABC at D. Given, ∠BAC = 55° and ∠ACB = 65°; calculate :

(i) ∠DCA,

(ii) ∠DAC,

(iii) ∠DCI,

(iv) ∠AIC.

Answer

(i) Join IA, IC and CD.

In △ABC,

⇒ ∠ABC + ∠BAC + ∠ACB = 180°

⇒ ∠ABC + 55° + 65° = 180°

⇒ ∠ABC + 120° = 180°

⇒ ∠ABC = 180° - 120° = 60°.

IB is the bisector of ∠ABC [As I is the incenter].

∠ABD = $\dfrac{1}{2}$∠ABC = $\dfrac{1}{2}$ x 60° = 30°.

We know that,

Angle in same segment are equal.

∴ ∠DCA = ∠ABD = 30°.

Hence, ∠DCA = 30°.

(ii) ∠CBD = ∠ABD = 30° [As IB is bisector of ∠ABC]

We know that,

Angle in same segment are equal.

∴ ∠DAC = ∠CBD = 30°.

Hence, ∠DAC = 30°.

(iii) As CI is the angle bisector of ∠ACB.

∠ACI = $\dfrac{1}{2}$∠ACB = $\dfrac{1}{2} \times 65°$ = 32.5°

From figure,

∠DCI = ∠DCA + ∠ACI = 30° + 32.5° = 62.5°

Hence, ∠DCI = 62.5°.

(iv) As AI is the angle bisector of ∠BAC.

∠IAC = $\dfrac{1}{2}$∠BAC = $\dfrac{1}{2} \times 55°$ = 27.5°

In △AIC,

⇒ ∠IAC + ∠ACI + ∠AIC = 180°

⇒ 27.5° + 32.5° + ∠AIC = 180°

⇒ 60° + ∠AIC = 180°

⇒ ∠AIC = 180° - 60° = 120°.

Hence, ∠AIC = 120°.

In the given figure, AB = AC = CD and ∠ADC = 38°. Calculate :

(i) Angle ABC

(ii) Angle BEC

Answer

Join BE.

(i) AC = CD

∠DAC = ∠ADC = 38° [Angles opposite to equal sides are equal]

In △ACD,

⇒ ∠DAC + ∠ADC + ∠ACD = 180°

⇒ 38° + 38° + ∠ACD = 180°

⇒ 76° + ∠ACD = 180°

⇒ ∠ACD = 180° - 76° = 104°

From figure,

⇒ ∠ACB + ∠ACD = 180° [BCD is a straight line]

⇒ ∠ACB + 104° = 180°

⇒ ∠ACB = 180° - 104° = 76°.

Given,

AB = AC

∴ ∠ABC = ∠ACB = 76°. [As angles opposite to equal sides are equal]

Hence, ∠ABC = 76°.

(ii) In △ABC,

⇒ ∠BAC + ∠ACB + ∠ABC = 180° [Angle sum property of triangle]

⇒ ∠BAC + 76° + 76° = 180°

⇒ ∠BAC + 152° = 180°

⇒ ∠BAC = 180° - 152° = 28°.

We know that,

Angles in same segment are equal.

⇒ ∠BEC = ∠BAC = 28°.

Hence, ∠BEC = 28°.

In the given figure, AC is the diameter of circle, centre O. Chord BD is perpendicular to AC. Write down the angles p, q and r in terms of x.

Answer

We know that,

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

⇒ ∠AOB = 2∠ACB

⇒ x = 2q

⇒ q = $\dfrac{x}{2}$.

Angles in the same segment are equal.

∴ ∠ADB = ∠ACB = q = $\dfrac{x}{2}$

From figure,

∠ADC = 90° [Angle in semi-circle is a right angle.]

∠BDC = ∠ADC - ∠ADB = 90° - $\dfrac{x}{2}$.

∴ r = 90° - $\dfrac{x}{2}$.

In △EBC,

⇒ ∠EBC + ∠CEB + ∠ECB = 180° [Angle sum property of triangle]

⇒ ∠EBC + 90° + q = 180°

⇒ ∠EBC = 90° - q

From figure,

∠DBC = ∠EBC = 90° - q = 90° - $\dfrac{x}{2}$.

Angles in the same segment are equal.

∴ ∠DAC = ∠DBC

⇒ p = 90° - $\dfrac{x}{2}$.

Hence, p = 90° - $\dfrac{x}{2}$, q = $\dfrac{x}{2}$ and r = 90° - $\dfrac{x}{2}$.

In the given figure, AOB is a diameter and DC is parallel to AB. If ∠CAB = x°; find (in terms of x) the values of :

(i) ∠COB,

(ii) ∠DOC,

(iii) ∠DAC,

(iv) ∠ADC.

Answer

(i) We know that,

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

⇒ ∠COB = 2∠CAB = 2x

Hence, ∠COB = 2x.

(ii) As DC || OB

⇒ ∠OCD = ∠COB = 2x [Alternate angles]

In △OCD,

OC = OD [Radius of same circle]

∠ODC = ∠OCD = 2x [Angles opposite to equal sides are equal]

⇒ ∠ODC + ∠OCD + ∠DOC = 180°

⇒ 2x + 2x + ∠DOC = 180°

⇒ ∠DOC = 180° - 4x.

Hence, ∠DOC = 180° - 4x.

(iii) We know that,

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

⇒ ∠DOC = 2∠DAC

⇒ ∠DAC = $\dfrac{1}{2}$∠DOC = $\dfrac{1}{2}$ x (180° - 4x) = 90° - 2x.

Hence, ∠DAC = 90° - 2x.

(iv) DC || AO

∴ ∠ACD = ∠OAC = x (Alternate angles are equal)

In △ADC,

⇒ ∠ADC + ∠DAC + ∠ACD = 180° [Angle sum property of triangle]

⇒ ∠ADC + 90° - 2x° + x = 180°

⇒ ∠ADC + 90° - x = 180°

⇒ ∠ADC = 180° - 90° + x

⇒ ∠ADC = 90° + x

Hence, ∠ADC = 90° + x.

In the given figure, PQ is the diameter of the circle whose center is O. Given, ∠ROS = 42°, calculate ∠RTS.

Answer

Join PS.

∠PSQ = 90° [Angle in semi-circle is a right angle.]

We know that,

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

⇒ ∠ROS = 2∠SPR

⇒ ∠SPR = $\dfrac{1}{2}$∠ROS = $\dfrac{42}{2}$ = 21°

From figure,

⇒ ∠SPT = ∠SPR = 21°.

From figure,

⇒ ∠PSQ = 90° [Angle in a semi-circle is a right angle.]

Since, QST is a straight line.

⇒ ∠PSQ + ∠PST = 180°

⇒ 90° + ∠PST = 180°

⇒ ∠PST = 90°.

In △PST,

⇒ ∠PTS + ∠PST + ∠SPT = 180° [Angle sum property of triangle]

⇒ ∠PTS + 90° + 21° = 180°

⇒ ∠PTS + 111° = 180°

⇒ ∠PTS = 180° - 111° = 69°.

From figure,

∠RTS = ∠PTS = 69°.

Hence, ∠RTS = 69°.

The given figure shows a circle with center O and ∠ABP = 42°. Calculate the measure of :

(i) ∠PQB

(ii) ∠QPB + ∠PBQ

Answer

Join AP.

(i) We know that,

Angle in a semi-circle is a right angle.

∠APB = 90°.

In △APB,

⇒ ∠APB + ∠ABP + ∠BAP = 180° [Angle sum property of triangle]

⇒ 90° + 42° + ∠BAP = 180°

⇒ ∠BAP + 132° = 180°

⇒ ∠BAP = 180° - 132° = 48°.

From figure,

∠PQB = ∠BAP = 48° [Angles in same segment are equal]

Hence, ∠PQB = 48°.

(ii) In △BQP,

⇒ ∠QPB + ∠PBQ + ∠PQB = 180° [Angle sum property of triangle]

⇒ ∠QPB + ∠PBQ + 48° = 180°

⇒ ∠QPB + ∠PBQ = 180° - 48°

⇒ ∠QPB + ∠PBQ = 132°.

Hence, ∠QPB + ∠PBQ = 132°.

In the given figure, M is the centre of the circle. Chords AB and CD are perpendicular to each other. If ∠MAD = x and ∠BAC = y :

(i) express ∠AMD in terms of x.

(ii) express ∠ABD in terms of y.

(iii) prove that : x = y.

Answer

Mark the point of intersection of AB and CD as L.

(i) In △AMD,

MA = MD [Radius of circle]

∴ ∠MAD = ∠MDA = x.

In △AMD,

⇒ ∠MAD + ∠MDA + ∠AMD = 180° [Angle sum property of triangle]

⇒ x + x + ∠AMD = 180°

⇒ ∠AMD = 180° - 2x.

Hence, ∠AMD = 180° - 2x.

(ii) Let the perpendicular chords AB and CD intersect each other at L.

From figure,

∠ALC = 90°.

In △ALC,

⇒ ∠LAC + ∠LCA + ∠ALC = 180° [Angle sum property of triangle]

⇒ y + ∠DCA + 90° = 180°

⇒ ∠DCA = 180° - 90° - y

⇒ ∠DCA = 90° - y

From figure,

∠ABD = ∠DCA [Angles in same segment are equal]

∴ ∠ABD = 90° - y.

Hence, ∠ABD = 90° - y.

(iii) 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.

∴ ∠AMD = 2∠ABD

∠ABD = $\dfrac{1}{2}$ ∠AMD = $\dfrac{1}{2} \times (180° - 2x)$ = 90° - x.

We have,

∠ABD = 90° - y and ∠ABD = 90° - x

⇒ 90° - y = 90° - x

⇒ x = y.

Hence, proved that x = y.