Circles

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 = $\frac{∠O}{2} = \frac{70°}{2}$ = 35°.

Hence, Option 2 is the correct option.

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 = $\frac{∠O}{2} = \frac{80°}{2}$ = 40°.

Hence, Option 3 is the correct option.

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.

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.

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.

(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°.

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°.

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°.

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 = $\frac{80}{2}$ = 40°.

Hence, c = 40°.

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°.

(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°.

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°.

Join AD.

We know that,

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

∠ADC = $\frac{1}{2}$∠AOC = $\frac{1}{2}$ 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°.

We know that,

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

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

(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°.

(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°.

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°.

(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°.

∠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.

(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 = $\frac{1}{2}$∠ABC = $\frac{1}{2}$ 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 = $\frac{1}{2}$∠ACB = $\frac{1}{2} \times 65°$ = 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 = $\frac{1}{2}$∠BAC = $\frac{1}{2} \times 55°$ = 27.5°

In △AIC,

⇒ ∠IAC + ∠ACI + ∠AIC = 180°

⇒ 27.5° + 32.5° + ∠AIC = 180°

⇒ 60° + ∠AIC = 180°

⇒ ∠AIC = 180° - 60° = 120°.

Hence, ∠AIC = 120°.

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°.

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 = $\frac{x}{2}$.

Angles in the same segment are equal.

∴ ∠ADB = ∠ACB = q = $\frac{x}{2}$

From figure,

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

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

∴ r = 90° - $\frac{x}{2}$.

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° - $\frac{x}{2}$.

Angles in the same segment are equal.

∴ ∠DAC = ∠DBC

⇒ p = 90° - $\frac{x}{2}$.

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

(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 = $\frac{1}{2}$∠DOC = $\frac{1}{2}$ 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.

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 = $\frac{1}{2}$∠ROS = $\frac{42}{2}$ = 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°.

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°.

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

We have,

∠ABD = 90° - y and ∠ABD = 90° - x

⇒ 90° - y = 90° - x

⇒ x = y.

Hence, proved that x = y.

Given,

AD || BC

Since,

Side BC is produced to point E.

We can say that,

AD || BE

From figure,

∠ADC = ∠DCE = 95° (Alternate angles are equal)

We know that,

The opposite angles of a cyclic quadrilateral are supplementary.

∴ ∠ADC + ∠CBA = 180°

⇒ ∠CBA = 180° - ∠ADC = 180° - 95° = 85°.

Hence, Option 1 is the correct option.

Since, ABC is an equilateral triangle.

∴ ∠ABC = 60°.

We know that,

The opposite angles of a cyclic quadrilateral are supplementary.

∴ ∠ABC + ∠ADC = 180°

⇒ 60° + ∠ADC = 180°

⇒ ∠ADC = 180° - 60° = 120°.

Hence, Option 4 is the correct option.

Join AC.

In △AOC,

Since,

OA = OC (Radius of same circle)

∴ ∠OAC = ∠OCA = x (let)

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + ∠AOC = 180°

⇒ ∠AOC = 180° - 2x

In △AOC,

By angle sum property of triangle,

⇒ ∠BAC + ∠ACB + ∠CBA = 180°

⇒ (30° + x) + (40° + x) + ∠CBA = 180°

⇒ ∠CBA + 70° + 2x = 180°

⇒ ∠CBA = 180° - 70° - 2x

⇒ ∠CBA = 110° - 2x

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.

∴ ∠AOC = 2∠CBA

⇒ 180° - 2x = 2(110° - 2x)

⇒ 180° - 2x = 220° - 4x

⇒ 4x - 2x = 220° - 180°

⇒ 2x = 40°

⇒ x = $\frac{40°}{2}$ = 20°.

⇒ ∠AOC = 180° - 2x

⇒ ∠AOC = 180° - 2(20°) = 180° - 40° = 140°.

Hence, Option 4 is the correct option.

Let ∠BPQ = x and ∠DQP = y

We know that,

The exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

From figure,

In cyclic quadrilateral APQC,

∠A = ∠DQP = y and ∠C = ∠BPQ = x

From figure,

APB is a straight line.

∴ ∠APQ + ∠BPQ = 180°

⇒ ∠APQ + x = 180°

⇒ ∠APQ = 180° - x

CQD is a straight line.

∴ ∠CQP + ∠DQP = 180°

⇒ ∠CQP + y = 180°

⇒ ∠CQP = 180° - y

In cyclic quadrilateral PQDB,

∠B = ∠CQP = 180° - y and ∠D = ∠APQ = 180° - x

⇒ ∠A + ∠B = y + (180° - y) = 180°

⇒ ∠C + ∠D = x + (180° - x) = 180°

We know that,

Sum of adjacent angles in a trapezium is 180°.

∴ ABDC is a trapezium.

∴ AC || BD.

Hence, Option 4 is the correct option.

We know that,

Sum of co-interior angles in a trapezium is equal to 180°.

⇒ ∠A + ∠D = 180°

⇒ 105° + ∠D = 180°

⇒ ∠D = 180° - 105° = 75°

We know that,

The opposite angles of a cyclic quadrilateral is 180°.

In cyclic quadrilateral ABCD,

⇒ ∠D + ∠B = 180°

⇒ 75° + ∠B = 180°

⇒ ∠B = 180° - 75° = 105°.

Hence, Option 1 is the correct option.

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 = $\frac{1}{2}$Reflex ∠AOB

= $\frac{1}{2}$ (360° - 140°)

= $\frac{1}{2} \times 220°$ = 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 = $\frac{40}{2}$ = 20°

Hence, ∠OAB = 20°.

(iv) We already found, ∠OBC = 60°.

⇒ ∠OBC = ∠CBA + ∠OBA

⇒ 60° = ∠CBA + 20°

⇒ ∠CBA = 60° - 20° = 40°

Hence, ∠CBA = 40°.

(i) We know,

Sum of angles in a triangle = 180°.

In △ABD,

⇒ ∠ADB + ∠ABD + ∠DAB = 180°

⇒ ∠ADB + 58° + 75° = 180°

⇒ ∠ADB + 133° = 180°

⇒ ∠ADB = 180° - 133° = 47°

From figure,

∠BDC = ∠ADC - ∠ADB = 77° - 47° = 30°.

Hence, ∠BDC = 30°.

(ii) We know that,

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

⇒ ∠BAD + ∠BCD = 180°

⇒ ∠BCD = 180° - 75° = 105°.

Hence, ∠BCD = 105°.

(iii) Join AC.

We know that,

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

∠BCA = ∠ADB = 47°

Hence, ∠BCA = 47°.

(i) We know that each angle in an equilateral triangle = 60°.

∴ ∠ACB = 60°

As angles in same segment are equal.

∴ ∠ADB = ∠ACB = 60°.

Hence, ∠ADB = 60°.

(ii) Join OA and OB.

We know that,

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

∴ ∠AOB = 2∠ACB = 2 x 60° = 120°.

∴ ∠AEB = $\frac{1}{2}$ Reflex ∠AOB

= $\frac{1}{2}$(360° - 120°) = $\frac{1}{2} \times 240°$

= 120°.

Hence, ∠AEB = 120°.

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°.

(i) ABCD is a cyclic quadrilateral.

So, ∠BAD + ∠BCD = 180° [As sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ ∠BCD = 180° - ∠BAD

= 180° - 96°

= 84°.

As DCE is a straight line.

∴ ∠BCE = 180° - ∠BCD

= 180° - 84° = 96°.

BCEF is a cyclic quadrilateral,

So, ∠BCE + ∠BFE = 180°

⇒ ∠BFE = 180° - ∠BCE

= 180° - 96°

= 84°.

Hence, ∠BCD = 84° and ∠BFE = 84°.

(ii) Now, ∠BAD + ∠BFE = 96° + 84° = 180°.

But these two are interior angles on the same side of a pair of lines AD and FE.

Hence, proved that AD || FE.

The figure of the parallelogram ABCD with a circle through its vertices A and B meeting side BC at point P and side AD at point Q is shown below:

We know that,

An exterior angle of a cyclic quadrilateral is equal to its opposite interior angle.

∴ ∠1 = ∠A ....... (i)

Also,

∠A = ∠C ...........(ii) [Opposite angles of a parallelogram are equal.]

From (i) and (ii) we get,

∠1 = ∠C .........(iii)

Also,

⇒ ∠C + ∠D = 180° [Sum of co-interior angles of a parallelogram = 180°]

⇒ ∠1 + ∠D = 180°

Hence, proved that PCDQ is cyclic.

(i) Let's assume that ABCD is a parallelogram which is inscribed in a circle.

⇒ ∠A = ∠C and ∠B = ∠D [Opposite angles of a parallelogram are equal]

⇒ ∠A + ∠C = 180° and ∠B + ∠D = 180° [Pair of opposite angles in a cyclic quadrilateral are supplementary]

⇒ 2∠A = 180° and 2∠B = 180°

⇒ ∠A = 90° and ∠B = 90°

⇒ ∠A = ∠C = 90° and ∠B = ∠D = 90°.

Also,

AB = CD and AD = BC [Opposite sides of parallelogram are equal]

Hence, proved that ABCD is a rectangle.

(ii) Let's assume that ABCD is a rhombus which is inscribed in a circle.

So, we have

∠BAD = ∠BCD [Opposite angles of a rhombus are equal]

From figure,

∠BAD + ∠BCD = 180° [Pair of opposite angles in a cyclic quadrilateral are supplementary]

⇒ 2∠BAD = 180°

Thus, ∠BAD = ∠BCD = 90°.

Also,

∠ABC = ∠ADC [Opposite angles of a rhombus are equal]

From figure,

∠ABC + ∠ADC = 180° [Pair of opposite angles in a cyclic quadrilateral are supplementary]

⇒ 2∠ABC = 180°

Thus, ∠ABC = ∠ADC = 90°.

Since, each side of rhombus are equal.

Thus,

Each angle of quadrilateral = 90° and all sides are equal.

Hence, proved that ABCD is a square.

Given, AB = AC

So, ∠B = ∠C ........(1) [As Angles opposite to equal sides are equal]

From figure,

DECB is a cyclic quadrilateral.

∴ ∠B + ∠DEC = 180° [Sum of opposite angles in cyclic quadrilateral = 180°]

⇒ ∠C + ∠DEC = 180° (Using 1)

But this is the sum of interior angles on one side of a transversal.

∴ DE || BC.

∴ ∠ADE = ∠B and ∠AED = ∠C [Corresponding angles]

Thus, ∠ADE = ∠AED

∴ AD = AE

⇒ AB - AD = AC - AE [As AB = AC]

⇒ BD = CE

Hence, we have DE || BC and BD = CE.

Hence, proved that DECB is an isosceles trapezium.

(i) We know that,

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

∴ ∠ABC = $\frac{1}{2}$ Reflex ∠COA

⇒ b = $\frac{1}{2}$ (360° - a)

⇒ 2b = 360° - a

⇒ a + 2b = 360° .....(1)

Hence, relationship between a and b is given by the equation : a + 2b = 360°.

(ii) From equation 1, we get :

⇒ a + 2b = 360°

⇒ a = 360° - 2b

As OABC is a parallelogram, the opposite angles are equal.

So, a = b

⇒ 360° - 2b = b

⇒ 3b = 360°

⇒ b = $\frac{360°}{3}$ = 120°

Let ∠OAB = x and ∠OCB = x.

⇒ ∠OAB + ∠OCB + ∠AOC + ∠ABC = 360°

⇒ x + x + a + b = 360°

⇒ 2x + 120° + 120° = 360°

⇒ 2x + 240° = 360°

⇒ 2x = 120°

⇒ x = $\frac{120°}{2}$ = 60°.

Hence, ∠OAB = 60°.

(i) Join RN and MS.

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

In ∆RMS

⇒ ∠RMS + ∠RSM + ∠MRS = 180° [Angle sum property of triangle]

⇒ 90° + ∠RSM + 29° = 180°

⇒ ∠RSM + 119° = 180°

⇒ ∠RSM = 180° - 119° = 61°.

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

⇒ ∠RNM + ∠RSM = 180°

⇒ ∠RNM + 61° = 180°

⇒ ∠RNM = 180° - 61° = 119°.

Hence, ∠RNM = 119°.

(ii) Now as RS || NM,

∠NMR = ∠MRS = 29° [Alternate angles are equal]

From figure,

∠NMS = ∠NMR + ∠RMS = 29° + 90° = 119°.

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

In quadrilateral NMSR,

⇒ ∠NRS + ∠NMS = 180°

⇒ ∠NRM + ∠MRS + ∠NMS = 180°

⇒ ∠NRM + 29° + 119° = 180°

⇒ ∠NRM = 180° - 119° - 29°

⇒ ∠NRM = 32°.

Hence, ∠NRM = 32°.

PQRS is a cyclic quadrilateral.

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

∴ ∠QRS + ∠QPS = 180° ...........(1)

Also,

∠QPS + ∠SPT = 180° [As QPT is a straight line] ........(2)

Subtracting equation (2) from (1) we get,

⇒ ∠QRS + ∠QPS - (∠QPS + ∠SPT) = 180° - 180°

⇒ ∠QRS - ∠SPT = 0

⇒ ∠QRS = ∠SPT ..........(3)

∠RQS = ∠RPS [Angles in same segment are equal] ........(4)

∠RPS = ∠SPT [As PS bisects ∠RPT] .......(5)

From (3), (4) and (5) we get :

⇒ ∠QRS = ∠RQS.

∴ SQ = SR [As sides opposite to equal angles are also equal]

Hence, proved that SQ = SR.

We know that,

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

⇒ Reflex ∠AOE = 2∠ADE

⇒ ∠ADE = $\frac{1}{2}$ Reflex ∠AOE

⇒ ∠ADE = $\frac{1}{2}$ (360° - 150°)

⇒ ∠ADE = $\frac{1}{2} \times 210°$ = 105°.

From figure,

⇒ ∠DAB + ∠BED = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°.]

⇒ ∠BED = 180° - ∠DAB = 180° - 51° = 129°.

Also,

⇒ ∠CEB + ∠BED = 180° [As CED is a straight line]

⇒ ∠CEB = 180° - ∠BED = 180° - 129° = 51°.

In △ADC,

⇒ ∠ADC + ∠ACD + ∠DAC = 180°

⇒ ∠ADE + ∠ACD + ∠DAO = 180° [From figure, ∠ADC = ∠ADE and ∠DAC = ∠DAO]

⇒ 105° + ∠ACD + 51° = 180°

⇒ ∠ACD = 180° - 105° - 51° = 24°.

From figure,

∠OCE = ∠ACD = 24°.

Hence, ∠CEB = 51° and ∠OCE = 24°.

We know that,

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

⇒ ∠APB = 2∠ACB

⇒ ∠ACB = $\frac{1}{2}$ ∠APB

⇒ ∠ACB = $\frac{1}{2} \times$ 150° = 75°.

From figure,

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

⇒ 75° + ∠BCD = 180°

⇒ ∠BCD = 180° - 75° = 105°.

Also,

⇒ Reflex ∠BQD = 2∠BCD [Angle at the center is double the angle at the circumference subtended by the same chord.]

⇒ (360° - x) = 2 x 105°

⇒ x = 360° - 210° = 150°.

Hence, x = 150°.

(i) We know that,

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

obtuse ∠AOB = 2∠APB = 2a°.

Hence, obtuse ∠AOB = 2a°.

(ii) OACB is a cyclic quadrilateral.

⇒ ∠AOB + ∠ACB = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°.]

⇒ ∠ACB + 2a° = 180°

⇒ ∠ACB = 180° - 2a°.

Hence, ∠ACB = 180° - 2a°.

(iii) Join AD and BD.

As, angles in same segment are equal.

∴ ∠ADB = ∠ACB = 180° - 2a°.

Hence, ∠ADB = 180° - 2a°.

We know that,

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

∠AOC = 2∠ABC = 2 × 55° = 110°.

∴ x = 110°.

ABCD is a cyclic quadrilateral.

∴ ∠ADC + ∠ABC = 180°

⇒ y + 55° = 180°

⇒ y = 180° - 55° = 125°.

Hence, x = 110° and y = 125°.

(i) We know that,

Angles in same segment are equal.

∴ ∠DAB = ∠BED = 65°.

Hence, ∠DAB = 65°.

(ii) We know that,

Angle in semi-circle is a right angle.

∴ ∠ADB = 90°.

In △ADB,

⇒ ∠ABD + ∠ADB + ∠DAB = 180°

⇒ ∠ABD + 90° + 65° = 180°

⇒ 155° + ∠ABD = 180°

⇒ ∠ABD = 180° - 155° = 25°.

As, AB || DC

∠BDC = ∠ABD = 25°.

Hence, ∠BDC = 25°.

(i) We know that,

Angle in semi-circle is a right angle.

∴ ∠AEB = 90°.

In △AEB,

⇒ ∠AEB + ∠EBA + ∠EAB = 180°

⇒ 90° + ∠EBA + 63° = 180°

⇒ 153° + ∠EBA = 180°

⇒ ∠EBA = 180° - 153° = 27°.

Hence, ∠EBA = 27°.

(ii) As, AB || ED

∴ ∠DEB = ∠EBA = 27° [Alternate angles]

BCDE is a cyclic quadrilateral.

∴ ∠DEB + ∠BCD = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°.]

⇒ 27° + ∠BCD = 180°

⇒ ∠BCD = 180° - 27° = 153°.

Hence, ∠BCD = 153°.

(i) ABCD is a cyclic quadrilateral.

∴ ∠DCB + ∠DAB = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°.]

⇒ ∠DAB = 180° - ∠DCB

⇒ ∠DAB = 180° - 120° = 60°.

Hence, ∠DAB = 60°.

(ii) We know that,

Angle in a semi-circle is a right angle.

∠ADB = 90°.

In △DAB,

⇒ ∠ADB + ∠DAB + ∠DBA = 180°

⇒ 90° + 60° + ∠DBA = 180°

⇒ 150° + ∠DBA = 180°

⇒ ∠DBA = 180° - 150° = 30°.

Hence, ∠DBA = 30°.

(iii) OD = OB (Radius of circle)

∴ ∠ODB = ∠OBD [Angles opposite to equal sides are equal]

From figure,

∠OBD = ∠DBA = 30°.

∴ ∠ODB = 30°.

As, DO || BC

∴ ∠DBC = ∠ODB = 30° [Alternate angles are equal.]

Hence, ∠DBC = 30°.

(iv) From figure,

∠ABC = ∠ABD + ∠DBC = 30° + 30° = 60°.

In cyclic quadrilateral ABCD,

∴ ∠ADC + ∠ABC = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°.]

⇒ ∠ADC = 180° - ∠ABC

⇒ ∠ADC = 180° - 60° = 120°.

Hence, ∠ADC = 120°.

In △AOD,

OA = OD [Radius of same circle]

∠AOD = ∠DAO [Angles opposite to equal sides are equal]

From figure,

⇒ ∠DAO = ∠DAB = 60°.

∴ ∠AOD = ∠DAO = ∠ADO = 60°

Hence, proved that △AOD is an equilateral triangle.

Let $\frac{x}{3} = \frac{y}{4} = \frac{z}{5}$ = k.

∴ x = 3k, y = 4k and z = 5k.

From figure,

∠BCP = ∠DCQ [Vertically opposite angles are equal.]

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

∠ABC = ∠BCP + ∠BPC = x + y = 3k + 4k = 7k

∠ADC = ∠DCQ + ∠DQC = x + z = 3k + 5k = 8k.

ABCD is a cyclic quadrilateral.

We know that,

⇒ ∠ABC + ∠ADC = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ 8k + 7k = 180°

⇒ 15k = 180°

⇒ k = $\frac{180}{15}$ = 12°.

x = 3k = 3 x 12° = 36°

y = 4k = 4 x 12° = 48°

z = 5k = 5 x 12° = 60°.

Hence, x = 36°, y = 48° and z = 60°.

(i) From figure,

⇒ ∠BOC + ∠BOA = 180° [As AOC is a straight line.]

⇒ ∠BOC + 80° = 180°

⇒ ∠BOC = 180° - 80° = 100°.

We know that,

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

⇒ ∠BOC = 2∠BEC

⇒ ∠BEC = $\frac{1}{2}$∠BOC = $\frac{100°}{2}$ = 50°.

Hence, ∠BEC = 50°.

(ii) Given,

DC || EB

∴ ∠DCE = ∠BEC = 50° [Alternate angles are equal]

We know that,

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

⇒ ∠AOB = 2∠ACB

⇒ ∠ACB = $\frac{1}{2}$∠AOB = $\frac{80°}{2}$ = 40°.

From figure,

∠BCD = ∠ACB + ∠ACE + ∠DCE = 40° + 10° + 50° = 100°.

Hence, ∠BCD = 100°.

(iii) As sum of opposite angles of cyclic quadrilateral = 180°.

⇒ ∠BED + ∠BCD = 180°

⇒ ∠BED = 180° - ∠BCD = 180° - 100° = 80°.

From figure,

⇒ ∠BED = ∠BEC + ∠CED

⇒ 80° = 50° + ∠CED

⇒ ∠CED = 80° - 50° = 30°.

Hence, ∠CED = 30°.

Join OA, OB, OC, OD.

In △OAB,

OA = OB [Radius of same circle]

∠1 = ∠2.

In △OBC,

OB = OC [Radius of same circle]

∠3 = ∠4.

In △OCD,

OC = OD [Radius of same circle]

∠5 = ∠6.

In △ODE,

OD = OE [Radius of same circle]

∠7 = ∠8.

In △OAB,

⇒ ∠1 + ∠2 + ∠a = 180° [By angle sum property of triangle] ..........(1)

In △OBC,

⇒ ∠3 + ∠4 + ∠b = 180° [By angle sum property of triangle] ..........(2)

In △OCD,

⇒ ∠5 + ∠6 + ∠c = 180° [By angle sum property of triangle] ..........(3)

In △ODE,

⇒ ∠7 + ∠8 + ∠d = 180° [By angle sum property of triangle] .........(4)

Adding (1), (2), (3) and (4) we get,

⇒ ∠1 + ∠2 + ∠a + ∠3 + ∠4 + ∠b + ∠5 + ∠6 + ∠c + ∠7 + ∠8 + ∠d + = 180° + 180° + 180° + 180°

⇒ ∠2 + ∠2 + ∠a + ∠3 + ∠3 + ∠b + ∠6 + ∠6 + ∠c + ∠7 + ∠7 + ∠d + = 720°

⇒ 2∠2 + 2∠3 + 2∠6 + 2∠7 + ∠a + ∠b + ∠c + ∠d = 720°

⇒ 2[∠2 + ∠3] + 2[∠6 + ∠7] + 180° = 720° [As ∠a + ∠b + ∠c + ∠d = 180°]

⇒ 2∠ABC + 2∠CDE = 540°

⇒ ∠ABC + ∠CDE = 270°.

Hence, ∠ABC + ∠CDE = 270°.

Join AB.

∠ABC = 90° [As angle in a semi-circle is a right angle.]

From figure,

∠ABE = ∠ABC - ∠CBE = 90° - 64° = 26°.

As angles in same segment are equal.

∠ACE = ∠ABE = 26°.

Given,

AC || ED

∴ ∠DEC = ∠ACE = 26° [Alternate angles are equal]

Hence, ∠DEC = 26°.

(i) In △ADP,

⇒ ∠PAD + ∠ADP + ∠DPA = 180° [Angle sum property of triangle]

⇒ ∠PAD + 85° + 40° = 180°

⇒ ∠PAD + 125° = 180°

⇒ ∠PAD = 180° - 125° = 55°.

From figure,

⇒ ∠BAD = ∠PAD = 55°.

Hence, ∠BAD = 55°.

(ii) We know that,

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

⇒ ∠ABC + ∠ADC = 180°

⇒ ∠ABC + 85° = 180°

⇒ ∠ABC = 180° - 85° = 95°.

In △AQB,

⇒ ∠AQB + ∠QAB + ∠ABQ = 180° [Angle sum property of triangle]

⇒ ∠AQB + ∠BAD + ∠ABC = 180° [From figure, ∠QAB = ∠BAD and ∠ABQ = ∠ABC]

⇒ ∠AQB + 55° + 95° = 180°

⇒ ∠AQB + 150° = 180°

⇒ ∠AQB = 180° - 150° = 30°.

From figure,

⇒ ∠DQB = ∠AQB = 30°.

Hence, ∠DQB = 30°.

Join PR.

(i) ∠PRQ = 90° [Angle in semi-circle is a right angle.]

In △PQR,

⇒ ∠RPQ + ∠PRQ + ∠PQR = 180° [Angle sum property of triangle]

⇒ ∠RPQ + 90° + 58° = 180°

⇒ ∠RPQ + 148° = 180°

⇒ ∠RPQ = 180° - 148° = 32°.

Hence, ∠RPQ = 32°.

(ii) As, SR || PQ,

∠PRS = ∠RPQ = 32° [Alternate angles are equal]

In cyclic quadrilateral PRST,

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

⇒ ∠STP = 180° - ∠PRS = 180° - 32° = 148°.

Hence, ∠STP = 148°.

Join BD.

(i) From figure,

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

In △OAD,

OA = OD [Radius of same circle]

∠OAD = ∠ODA = x (let) [As angles opposite to equal side are equal]

In △OAD,

⇒ ∠OAD + ∠ODA + ∠AOD = 180° [Angle sum property of triangle]

⇒ x + x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\frac{120°}{2}$ = 60°.

From figure,

∠ODB = ∠BDA - ∠ADO = 90° - 60° = 30°.

Given, OD || BC

∠DBC = ∠ODB = 30° [Alternate angles are equal]

Hence, ∠DBC = 30°.

(ii) We know that,

Angle in a semi-circle is a right angle.

∴ ∠BDA = 90°.

Since,

∠OAD = ∠ODA = ∠AOD = 60°.

From figure,

⇒ ∠BDA = ∠ODA + ∠ODB

⇒ 90° = 60° + ∠ODB

⇒ ∠ODB = 90° - 60° = 30°.

Given,

OD || BC

∴ ∠DBC = ∠ODB = 30° [Alternate angles are equal].

Hence, ∠DBC = 30°.

(iii) From figure,

∠ABC = ∠ABD + ∠DBC = 30° + 30° = 60°.

Since, sum of opposite nagles in a cyclic quadrilateral = 180°.

∴ ∠ABC + ∠ADC = 180°

⇒ 60° + ∠ADC = 180°

⇒ ∠ADC = 120°.

Hence, ∠ADC = 120°.

Join AB and AD.

(i) 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∠APB = 2 × 75° = 150°.

Hence, ∠AOB = 150°.

(ii) We know that,

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

∴ ∠ACB + ∠AOB = 180°

⇒ ∠ACB + 150° = 180°

⇒ ∠ACB = 180° - 150° = 30°.

Hence, ∠ACB = 30°.

(iii) From figure,

∠ACD = ∠ACB + ∠BCD = 30° + 40° = 70°.

We know that,

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

In cyclic quadrilateral ABDC,

∴ ∠ABD + ∠ACD = 180°

⇒ ∠ABD + 70° = 180°

⇒ ∠ABD = 180° - 70° = 110°.

Hence, ∠ABD = 110°.

(iv) We know that,

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

In cyclic quadrilateral AOBD,

∴ ∠ADB + ∠AOB = 180°

⇒ ∠ADB + 150° = 180°

⇒ ∠ADB = 180° - 150° = 30°.

Hence, ∠ADB = 30°.

(i) We know that,

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

In cyclic quadrilateral ABCD,

∴ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 65° = 180°

⇒ ∠BCD = 180° - 65° = 115°.

Hence, ∠BCD = 115°.

(ii) In △ABD,

⇒ ∠ADB + ∠BAD + ∠DBA = 180° [Angle sum property of triangle]

⇒ ∠ADB + 65° + 70° = 180°

⇒ ∠ADB + 135° = 180°

⇒ ∠ADB = 180° - 135° = 45°.

We know that,

Angles in same segment are equal.

∴ ∠ACB = ∠ADB = 45°.

Hence, ∠ADB = 45°.

From figure,

∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°.

Since, angle in a semi-circle is a right angle.

Hence, proved that AC is a diameter.

Given,

Chord AB : Chord CD = 3 : 5

∴ ∠AOB : ∠COD = 3 : 5

⇒ 60° : ∠COD = 3 : 5

⇒ $\frac{60°}{∠COD} = \frac{3}{5}$

⇒ ∠COD = $\frac{60° \times 5}{3} = \frac{300°}{3}$ = 100°.

Hence, Option 4 is the correct option.

From figure,

In △OAB,

OA = OB (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

∴ ∠OBA = ∠OAB = 55°

By angle sum property of triangle,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 55° + 55° + ∠AOB = 180°

⇒ ∠AOB + 110° = 180°

⇒ ∠AOB = 180° - 110° = 70°.

We know that,

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

∴ ∠AOB = 2∠ACB

∠ACB = $\frac{∠AOB}{2} = \frac{70°}{2}$ = 35°.

Hence, Option 2 is the correct option.

Join OD.

Since, AB is the side of square.

∴ ∠AOB = $\frac{360°}{4}$ = 90°.

Since, BC is the side of regular hexagon.

∴ ∠BOC = $\frac{360°}{6}$ = 60°.

We know that,

Equal arcs subtends equal angles at the center.

Since, arc AD = arc CD

∴ ∠AOD = ∠COD = x (let)

From figure,

⇒ ∠AOD + ∠COD + ∠AOB + ∠BOC = 360°

⇒ x + x + 90° + 60° = 360°

⇒ 2x + 150° = 360°

⇒ 2x = 360° - 150°

⇒ 2x = 210°

⇒ x = $\frac{210°}{2}$ = 105°.

Hence, Option 2 is the correct option.

Join OA and OB.

Given,

AB is side of a regular pentagon.

∴ ∠AOB = $\frac{360°}{5}$ = 72°.

We know that,

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

∴ ∠AOB = 2∠ACB

∠ACB = $\frac{∠AOB}{2} = \frac{72°}{2}$ = 36°.

Hence, Option 1 is the correct option.

Given,

Chords AB, CD and EF are equal.

∴ ∠AOB = ∠COD = ∠EOF = x (let)

Chords BC, DE and FA are equal.

∴ ∠BOC = ∠DOE = ∠AOF = y (let)

From figure,

⇒ ∠AOB + ∠COD + ∠EOF + ∠BOC + ∠DOE + ∠AOF = 360°

⇒ x + x + x + y + y + y = 360°

⇒ 3x + 3y = 360°

⇒ 3(x + y) = 360°

⇒ x + y = $\frac{360°}{3}$

⇒ x + y = 120°

From figure,

∠AOC = ∠AOB + ∠BOC = x + y = 120°.

Hence, Option 4 is the correct option.

Join AE, OB and OC.

(i) As AOD is the diameter

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

But, given ∠DEF = 110°

So,

∠AEF = ∠DEF - ∠AED = 110° - 90° = 20°.

Hence, ∠AEF = 20°.

(ii) Also given, Chord AB = Chord BC = Chord CD

So,

∠AOB = ∠BOC = ∠COD [Equal chords subtends equal angles at the centre]

From figure,

⇒ ∠AOB + ∠BOC + ∠COD = 180° [AOD is a straight line]

⇒ ∠AOB = ∠BOC = ∠COD = $\frac{180°}{3}$ = 60°

Now, in ∆OAB we have

OA = OB [Radii of same circle]

So, ∠OAB = ∠OBA [Angles opposite to equal sides are equal]

In ∆OAB,

⇒ ∠OAB + ∠OBA + ∠AOB = 180° [By angle sum property of triangle]

⇒ ∠OAB + ∠OBA + 60° = 180°

⇒ ∠OAB + ∠OBA = 180° - 60° = 120°.

Since, ∠OAB = ∠OBA

∴ ∠OAB = ∠OBA = $\frac{120°}{2}$ = 60°.

Now, in cyclic quadrilateral ADEF

⇒ ∠DEF + ∠DAF = 180° [As sum of opposite angles in cyclic quadrilateral = 180°]

⇒ ∠DAF = 180° - ∠DEF

⇒ ∠DAF = 180° - 110° = 70°.

From figure,

∠FAB = ∠DAF + ∠OAB = 70° + 60° = 130°.

Hence, ∠FAB = 130°.

Join OP, OQ, OR and OS.

Given, PQ = QR = RS

∴ ∠POQ = ∠QOR = ∠ROS = x (let) [Equal chords subtends equal angles at the centre]

Arc PQRS subtends ∠POS at the centre and ∠PTS at the circumference of the circle.

Thus,

∠POS = 2 x ∠PTS = 2 x 75° = 150° [As angle subtended at the centre by the arc is double that it subtends at any point on the circumference of the circle.]

⇒ ∠POQ + ∠QOR + ∠ROS = 150°

⇒ x + x + x = 150°

⇒ 3x = 150°

⇒ x = $\frac{150°}{3}$ = 50°.

In ∆OPQ we have,

⇒ OP = OQ [Radii of the same circle]

⇒ ∠OPQ = ∠OQP = y (let) [Angles opposite to equal sides are equal]

In ∆OPQ,

⇒ ∠OPQ + ∠OQP + ∠POQ = 180°

⇒ ∠OPQ + ∠OQP + 50° = 180°

⇒ ∠OPQ + ∠OQP = 180° - 50°

⇒ ∠OPQ + ∠OQP = 130°

⇒ 2y = 130°

⇒ y = $\frac{130°}{2}$ = 65°

⇒ ∠OPQ = ∠OQP = y = 65°.

In ∆OQR we have,

⇒ OQ = OR [Radii of the same circle]

⇒ ∠OQR = ∠ORQ = z (let) [Angles opposite to equal sides are equal]

In ∆OQR

⇒ ∠OQR + ∠ORQ + ∠QOR = 180°

⇒ z + z + 50° = 180°

⇒ 2z = 180° - 50°

⇒ 2z = 130°

⇒ z = $\frac{130°}{2}$ = 65°

⇒ ∠OQR = ∠ORQ = z = 65°.

(i) Hence, ∠POS = 150°.

(ii) Hence, ∠QOR = 50°.

(iii) From figure,

∠PQR = ∠PQO + ∠OQR = 65° + 65° = 130°.

Hence, ∠PQR = 130°.

Join OC.

(i) We know that,

Each side of a regular hexagon, inscribed in a circle subtends an angle of 60° at the centre.

⇒ ∠AOB = 60°.

Hence, ∠AOB = 60°.

(ii) We know that,

Angle at the centre is twice the angle at remaining circumference.

∴ ∠AOB = 2∠ACB

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

Hence, ∠ACB = 30°.

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

∠AOC = $\frac{360°}{8}$ = 45°.

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

∴ ∠AOC = 2∠ABC

∴ ∠ABC = $\frac{1}{2}$ x ∠AOC = $\frac{1}{2}$ x 45° = 22.5°

Hence, ∠ABC = 22.5°.

(i) Join EB and EC.

In cyclic quadrilateral ABCE,

⇒ ∠ABC + ∠AEC = 180° [Sum of opposite angles in cyclic quadrilateral = 180°]

⇒ 132° + ∠AEC = 180°

⇒ ∠AEC = 180° - 132° = 48°.

Since, AB = BC.

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.

∴ ∠AEB = $\frac{1}{2}$∠AEC

= $\frac{1}{2} \times 48°$ = 24°.

Hence, ∠AEB = 24°.

(ii) We know that,

Equal chords subtend equal angles at the circumference of the circle.

∠AEB = ∠BEC = ∠CED = 24°

∠AED = ∠AEB + ∠BEC + ∠CED = 24° + 24° + 24° = 72°.

Hence, ∠AED = 72°.

(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.

∴ ∠COD = 2∠CED = 2 × 24° = 48°.

Hence, ∠COD = 48°.

Join AD and BD.

(i) Given,

⇒ arc AB = 2 arc BC

⇒ ∠AOB = 2∠BOC

⇒ ∠BOC = $\frac{1}{2}$∠AOB = $\frac{1}{2} \times 108°$ = 54°.

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.

∴ ∠BOC = 2∠CAB

⇒ ∠CAB = $\frac{1}{2}$∠BOC = $\frac{1}{2} \times 54°$ = 27°.

Hence, ∠CAB = 27°.

(ii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∴ ∠AOB = 2∠ACB

∠ACB = $\frac{1}{2}$∠AOB = $\frac{1}{2} \times 108°$ = 54°.

In cyclic quadrilateral ADBC,

⇒ ∠ADB + ∠ACB = 180° [As sum of opposite angles in cyclic quadrilateral = 180°]

⇒ ∠ADB + 54° = 180°

⇒ ∠ADB = 180° - 54° = 126°.

Hence, ∠ADB = 126°.

Join OA, OB and OC.

Since, AB is the side of regular pentagon,

∠AOB = $\frac{360°}{5}$ = 72°.

Since, AC is the side of regular hexagon,

∠AOC = $\frac{360°}{6}$ = 60°.

From figure,

⇒ ∠AOB + ∠AOC + reflex∠BOC = 360°

⇒ 72° + 60° + reflex∠BOC = 360°

⇒ reflex∠BOC + 132° = 360°

⇒ reflex∠BOC = 360° - 132° = 228°.

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.

Arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.

∴ ∠BOC = 2∠BAC

⇒ ∠BAC = $\frac{1}{2}∠BOC = \frac{1}{2} \times 228°$ = 114°.

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

∴ ∠AOC = 2∠ABC

⇒ ∠ABC = $\frac{1}{2} \times ∠AOC = \frac{1}{2} \times 60°$ = 30°.

Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∴ ∠AOB = 2∠ACB

⇒ ∠ACB = $\frac{1}{2} \times ∠AOB = \frac{1}{2} \times 72°$ = 36°.

Hence, angles of triangle are ∠ABC = 30°, ∠ACB = 36° and ∠BAC = 114°.

Let O be the center of the circle.

Join BC, BO, CO and EO.

Since, BD is the side of a regular hexagon,

∴ ∠BOD = $\frac{360°}{6}$ = 60°.

Since, DC is the side of a regular pentagon,

∴ ∠COD = $\frac{360°}{5}$ = 72°.

In △BOD,

OB = OD [Radii of same circle]

∴ ∠OBD = ∠ODB = x (let) [Angles opposite to equal sides are equal]

⇒ ∠OBD + ∠ODB + ∠BOD = 180°

⇒ x + x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\frac{120°}{2}$ = 60°.

(i) In △OCD,

⇒ OD = OC [Radii of same circle]

⇒ ∠ODC = ∠OCD = y (let) [Angles opposite to equal sides are equal]

⇒ ∠OCD + ∠ODC + ∠COD = 180°

⇒ y + y + 72° = 180°

⇒ 2y = 180° - 72°

⇒ 2y = 108°

⇒ y = $\frac{108°}{2}$ = 54°.

From figure,

∠ADC = ∠ODC = y = 54°.

Hence, ∠ADC = 54°.

(ii) From figure,

∠BDA = ∠BDO = 60°.

Hence, ∠BDA = 60°.

(iii) We know that,

Angle at the centre is twice the angle at remaining circumference.

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

∴ ∠AOC = 2∠ABC

⇒ ∠ABC = $\frac{1}{2}$∠AOC

⇒ ∠ABC = $\frac{1}{2}$[∠AOD - ∠COD]

⇒ ∠ABC = $\frac{1}{2}$[180° - 72°]

⇒ ∠ABC = $\frac{1}{2} \times 108°$

⇒ ∠ABC = 54°.

Hence, ∠ABC = 54°.

(iv) In cyclic quadrilateral AECD,

⇒ ∠AEC + ∠ADC = 180°

⇒ ∠AEC + 54° = 180°

⇒ ∠AEC = 180° - 54°

⇒ ∠AEC = 126°.

Hence, ∠AEC = 126°.

We know that,

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

In cyclic quadrilateral ABCD,

∠C = x°, ∠B = p°, ∠D = y° and ∠A = z°.

∴ x° + y° + z° + p° = ∠C + ∠D + ∠A + ∠B

We know that,

Sum of angles in a cyclic quadrilateral is 360°.

∴ ∠A + ∠B + ∠C + ∠D = 360°.

Hence, Option 3 is the correct option.

Join OB.

From figure,

OB = OA (Radius of same circle) ........(1)

We know that,

Sides of rhombus are equal.

∴ OA = AB .............(2)

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

⇒ OA = OB = AB

∴ OAB is an equilateral triangle.

Since, diagonals of rhombus bisect the interior angles.

In △OAB,

∠AOB = $\frac{x}{2}$

∠OBA = $\frac{y}{2}$

Since, each angle of equilateral triangle is 60°.

∴ ∠AOB = 60°

⇒ $\frac{x}{2} = 60°$

⇒ x = 120°.

∴ ∠OBA = 60°

⇒ $\frac{y}{2} = 60°$

⇒ y = 120°.

Substituting value of x and y in L.H.S. of equation x° + 2y° = 360°, we get :

⇒ 120° + 2(120°)

⇒ 120° + 240°

⇒ 360°.

Since, L.H.S. = R.H.S.

Hence, Option 3 is the correct option.

Given,

Arcs AB and BC are of lengths in the ratio 11 : 4.

∴ ∠AOB : ∠BOC = 11 : 4

⇒ ∠AOB : 32° = 11 : 4

$\Rightarrow \frac{∠AOB}{32°} = \frac{11}{4}$

⇒ ∠AOB = $\frac{11}{4} \times 32°$ = 88°.

Hence, Option 2 is the correct option.

Since,

AB is the side of regular pentagon.

∴ ∠AOB = $\frac{360°}{5}$ = 72°.

BC is the side of regular hexagon.

∴ ∠BOC = $\frac{360°}{6}$ = 60°.

From figure,

∠AOC = ∠AOB + ∠BOC = 72° + 60° = 132°.

We know that,

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

⇒ ∠AOC = 2∠APC

⇒ ∠APC = $\frac{∠AOC}{2} = \frac{132°}{2}$ = 66°.

Hence, Option 2 is the correct option.

Given,

Chord BC = chord CD

∴ ∠BOC = ∠COD = x (let)

From figure,

∠BOD = ∠BOC + ∠COD = x + x = 2x

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.

⇒ ∠BOD = 2∠BAD

⇒ 2x = 2 × 80°

⇒ x = 80°.

Hence, Option 2 is the correct option.

We know that,

Angle in semicircle is a right angle.

If AC is the diameter, then ∠ADC = ∠ABC = 90°.

∴ Assertion (A) is true.

From figure,

In △ADB,

By angle sum property of triangle,

∴ ∠ADB + ∠DBA + ∠BAD = 180°

⇒ ∠ADB + 40° + 95° = 180°

⇒ ∠ADB + 135° = 180°

⇒ ∠ADB = 180° - 135° = 45°.

From figure,

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

∴ Reason (R) is true.

Hence, Option 3 is the correct option.

Since BD and AC are diameters, that means:

∠ABC and ∠BCD are right angles (Angles in a semicircle is a right angle)

In △ DBC, using angle sum property,

⇒ ∠DBC + ∠BCD + ∠BDC = 180°

⇒ 50° + 90° + ∠BDC = 180°

⇒ 140° + ∠BDC = 180°

⇒ ∠BDC = 180° - 140°

⇒ ∠BDC = 40°

We know that, angles in the same segment of a circle are equal.

⇒ ∠BAC = ∠BDC

⇒ ∠BAC = 40°

So, assertion and reason are true and reason clearly explains assertion.

Hence, option 3 is the correct option.

Join AC.

Since AB is diameter, that means :

∠ACB = 90° (Angles in a semicircle is a right angle)

In △ ABC, using angle sum property,

⇒ ∠ABC + ∠ACB + ∠BAC = 180° ....................(1)

⇒ 60° + 90° + ∠BAC = 180°

⇒ 150° + ∠BAC = 180°

⇒ ∠BAC = 180° - 150°

⇒ ∠BAC = 30°

So, assertion (A) is false.

From equation (1),

⇒ ∠ABC + 90° + ∠BAC = 180°

⇒ ∠ABC + ∠BAC = 180° - 90°

⇒ ∠ABC + ∠BAC = 90°

So, reason (R) is true.

Hence, option 2 is the correct option.

Join DB and CB.

It is given AB is diameter of the circle and angles in a semicircle is a right angle.

⇒ ∠ACB = 90°

⇒ ∠ACD + ∠DCB = 90°

⇒ 38° + ∠DCB = 90°

⇒ ∠DCB = 90° - 38°

⇒ ∠DCB = 52°

We know that, angles subtended by the same chord in the same segment of a circle are equal.

⇒ ∠BAD (x) = ∠BCD = 52°

So, assertion (A) is false but reason (R) is true.

Hence, option 2 is the correct option.

In Δ APD and Δ BPC,

⇒ ∠APD = ∠BPC (Vertically opposite angles are equal)

⇒ ∠ADP = ∠BCP (Angles in same segment are equal)

∴ Δ APD ∼ Δ BPC (By A.A. similarity)

Corresponding sides of similar triangles are proportional.

$\Rightarrow \frac{AP}{PB} = \frac{PD}{PC}$ ........(1)

So, reason (R) is true.

Solving (1),

⇒ AP x PC = PD x PB

So, assertion (A) is true and R is the correct reason for A.

Hence, option 3 is the correct option.

We know that, the angle, subtended by an arc of a circle is double the angle subtended by it at any point on the remaining part of the circle.

So, statement 2 is true.

⇒ ∠AOC = 2x

⇒ 160° = 2x

⇒ x = $\frac{160°}{2}$ = 80°

We know that, sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

⇒ x + y = 180°

⇒ 80° + y = 180°

⇒ y = 180° - 80° = 100°

So, statement 1 is false.

Hence, option 4 is the correct option.

It is given that AC is diameter and angles in a semicircle is a right angle.

⇒ ∠ABC = 90°

Since, AE is parallel to BC and AB is transversal.

⇒ ∠ABC + ∠BAE = 180° [The sum of co-interior angles formed when a transversal intersects two parallel lines is always 180°]

⇒ 90° + ∠BAE = 180°

⇒ ∠BAE = 180° - 90°

⇒ ∠BAE = 90°

⇒ ∠BAC + ∠EAC = 90°

⇒ 50° + ∠EAC = 90°

⇒ ∠EAC = 90° - 50°

⇒ ∠EAC = 40°

AEDC form a cyclic quadrilateral and sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

⇒ ∠EDC + ∠EAC = 180°

⇒ ∠EDC + 40° = 180°

So, statement 1 is false but statement 2 is true.

Hence, option 4 is the correct option.

Given,

⇒ OB = OC

⇒ ∠BOC = ∠BCO = 20° (Angles opposite to equal sides of a triangle are always equal)

In △ OBC, using angle sum property,

⇒ ∠OBC + ∠BCO + ∠BOC = 180°

⇒ ∠OBC + 20° + 20° = 180°

⇒ ∠OBC + 40° = 180°

⇒ ∠OBC = 180° - 40°

⇒ ∠OBC = 140°

∠OBC and ∠OBA forms linear pairs of angle.

⇒ ∠OBC + ∠OBA = 180°

⇒ 140° + ∠OBA = 180°

⇒ ∠OBA = 180° - 140°

⇒ ∠OBA = 40°

Since OB = OA (Radii of same circle)

⇒ ∠OBA = ∠OAB = 40° (Angles opposite to equal sides of a triangle are always equal)

Using exterior angle property, the exterior angle of a triangle is equal to the sum of the two opposite interior angles.

In triangle OAC,

⇒ ∠EOA = ∠OAC + ∠OCA

⇒ x = ∠OAB + 20°

⇒ x = ∠OBA + 20° = 40° + 20° = 60°

So, statement 1 is false and statement 2 is true.

Hence, option 4 is the correct option.

Since, the angle, which an arc of a circle subtends at the center of the circle is double the angle which it subtends at any point on the remaining part of the circumference.

⇒ ∠AOC = 2 x ∠ADC

⇒ 120° = 2 x ∠ADC

⇒ ∠ADC = $\frac{120°}{2}$ = 60°

ABCD form a cyclic quadrilateral and sum of opposite angles of cyclic quadrilateral is 180°.

⇒ ∠ADC + ∠ABC = 180°

⇒ 60° + ∠ABC = 180°

⇒ ∠ABC = 180° - 60°

⇒ ∠ABC = 120°

So, both statement are true.

Hence, option 1 is the correct option.

As angles in same segment are equal.

∴ ∠ABD = ∠ACD = 30°.

From figure,

∠ADB = 90° [As angle in semi-circle is a right angle.]

In △ADB,

⇒ ∠ABD + ∠ADB + ∠BAD = 180° [By angle sum property of triangle]

⇒ 30° + 90° + x = 180°

⇒ 120° + x = 180°

⇒ x = 180° - 120° = 60°.

Hence, the value of x = 60°.

Join OB and OC.

We know that,

Angle at the centre is twice the angle at remaining circumference.

∴ ∠BOC = 2∠BAC = 2 × 30° = 60°

In △OBC,

⇒ OB = OC [Radii of same circle]

⇒ ∠OBC = ∠OCB = x (let) [As angles opposite to equal sides are equal]

⇒ ∠BOC + ∠OBC + ∠OCB = 180°

⇒ 60° + x + x = 180°

⇒ 2x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\frac{120°}{2}$ = 60°.

∴ ∠OBC = ∠OCB = ∠BOC = 60°.

Hence, △OBC is an equilateral triangle.

∴ OB = OC = BC.

Hence, proved that BC is equal to the radius of the circumcircle of the triangle ABC.

Join OE and AB.

Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.

We know that,

Angle at the centre is twice the angle at remaining circumference.

∴ ∠EOC = 2∠EBC = 2 x 65° = 130°.

Now, in ∆OEC

OE = OC [Radii of the same circle]

So, ∠OEC = ∠OCE [Angle opposite to equal sides are equal.]

In ∆OCE by angle sum property,

⇒ ∠OEC + ∠OCE + ∠EOC = 180°

⇒ 2∠OCE + 130° = 180°

⇒ 2∠OCE = 180° - 130°

⇒ 2∠OCE = 50°

⇒ ∠OCE = $\frac{50°}{2}$ = 25°.

Given, AC || ED

∴ ∠DEC = ∠OCE [Alternate angles are equal]

⇒ ∠DEC = 25°.

Hence, ∠DEC = 25°.

(i) Given that BD is a diameter of the circle.

We know that,

Angle in a semicircle is a right angle.

So, ∠BCD = 90°

Also given that,

∠DBC = 58°

In ∆BDC,

⇒ ∠DBC + ∠BCD + ∠BDC = 180° [Angle sum property of triangle]

⇒ 58° + 90° + ∠BDC = 180°

⇒ 148° + ∠BDC = 180°

⇒ ∠BDC = 180° - 148° = 32°.

Hence, ∠BDC = 32°.

(ii) We know that, the opposite angles of a cyclic quadrilateral are supplementary.

So, in cyclic quadrilateral BECD

⇒ ∠BEC + ∠BDC = 180°

⇒ ∠BEC + 32° = 180°

⇒ ∠BEC = 180° - 32° = 148°

Hence, ∠BEC = 148°.

(iii) In cyclic quadrilateral ABEC,

⇒ ∠BAC + ∠BEC = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

⇒ ∠BAC + 148° = 180°

⇒ ∠BAC = 180° - 148°= 32°.

Hence, ∠BAC = 32°.

Given,

In cyclic quad. ABCD

AF || CB and DA is produced to E such that ∠ADC = 92° and ∠FAE = 20°.

From figure,

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

⇒ ∠B + 92° = 180°

⇒ ∠B = 180° - 92° = 88°

As AF || CB,

∠FAB = ∠B = 88° [Alternate angles are equal]

But, ∠FAE = 20° [Given]

From figure,

∠BAE = ∠BAF + ∠FAE = 88° + 20° = 108°.

∠BAD = 180° - ∠BAE = 180° - 108° = 72°.

∠BCD + ∠BAD = 180° [As sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ ∠BCD + 72° = 180°

⇒ ∠BCD = 180° - 72° = 108°.

Hence, ∠BCD = 108°.

Join DB and DC, IB and IC.

(i) From figure,

∠DAC = $\frac{1}{2}$∠BAC [As I is the incenter]

∠DAC = $\frac{1}{2}$ x 66° = 33°.

As angle in same segment are equal.

∴ ∠DBC = ∠DAC = 33°.

Hence, ∠DBC = 33°.

(ii) Since, I is the incentre of ∆ABC, IB bisects ∠ABC.

∴ ∠IBC = $\frac{1}{2}$∠ABC

= $\frac{1}{2} \times 80°$ = 40°.

Hence, ∠IBC = 40°.

(iii) In ∆ABC,

⇒ ∠ACB + ∠ABC + ∠BAC = 180° [By angle sum property]

⇒ ∠ACB = 180° - ∠ABC - ∠BAC

⇒ ∠ACB = 180° - 80° - 66°

⇒ ∠ACB = 180° - 146°

⇒ ∠ACB = 34°.

As I is incenter so, IC bisects ∠C

∴ ∠ICB = $\frac{1}{2}$∠ACB = $\frac{1}{2} \times 34°$ = 17°.

In ∆IBC

⇒ ∠IBC + ∠ICB + ∠BIC = 180° [By angle sum property of triangle]

⇒ 40° + 17° + ∠BIC = 180°

⇒ 57° + ∠BIC = 180°

⇒ ∠BIC = 180° - 57° = 123°.

Hence, ∠BIC = 123°.

(i) Join AC and BD.

∠DAC = ∠DBC = x° [Angles in the same segment are equal]

∠DCA = ∠DAC = x° [As angles opposite to equal sides are equal]

Also, we have

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

Hence, ∠ABD = x°.

(ii) In ∆ABP

⇒ Ext. ∠ABC = ∠BAP + ∠APB

But, ∠BAP = ∠APB [As angles opposite to equal sides are equal]

⇒ 2x° = ∠APB + ∠APB = 2∠APB

⇒ 2∠APB = 2x°

⇒ ∠APB = x°

Hence, ∠APB = x°.

Thus, ∠APB = ∠DBC = x° [These are corresponding angles]

Hence, proved that AP || DB.

Join EB.

We know that,

In cyclic quadrilateral sum of opposite angles = 180°.

In cyclic quad. ABEP

⇒ ∠APE + ∠ABE = 180° .....(1) [Opposite angles of a cyclic quad. are supplementary]

Similarly, in cyclic quad. BCQE

⇒ ∠CQE + ∠CBE = 180° .....(2)

Adding (1) and (2), we have

⇒ ∠APE + ∠ABE + ∠CQE + ∠CBE = 180° + 180°

⇒ ∠APE + ∠ABE + ∠CQE + ∠CBE = 360° .....(3)

From figure,

∠ABE + ∠CBE = 180° [Linear pair]

Putting this value of ∠ABE + ∠CBE in Eq 3 we get,

∠APE + ∠CQE + 180° = 360°

⇒ ∠APE + ∠CQE = 360° - 180°

⇒ ∠APE + ∠CQE = 180°

Hence, proved that ∠APE and ∠CQE are supplementary.

We know that,

Angle at the centre is twice the angle at remaining circumference.

Arc AC subtends ∠AOC at the centre and ∠ADC at the remaining part of the circle.

⇒ ∠AOC = 2∠ADC

⇒ ∠AOC = 2 x 32° = 64°

From figure,

∠AOC and ∠BOC are linear pair,

∴ ∠AOC + ∠BOC = 180°

⇒ 64° + ∠BOC = 180°

⇒ ∠BOC = 180° - 64° = 116°.

Hence, ∠BOC = 116°.

AF is the angle bisector of ∠A,

∴ ∠BAF = ∠DAF

Also,

∠DAE = ∠BAE .......(1)

and

∠DAE = ∠AEB ........(2) [Alternate angles are equal]

From (1) and (2) we get,

∠BAE = ∠AEB

(i) In △ABE,

⇒ ∠BAE + ∠AEB + ∠ABE = 180°

⇒ ∠AEB + ∠AEB + ∠ABE = 180°

⇒ 2∠AEB + ∠ABE = 180°

⇒ ∠ABE = 180° - 2∠AEB

Since, ABCD is a cyclic quadrilateral and sum of opposite angles in a cyclic quadrilateral = 180°.

∴ ∠ABC + ∠ADC = 180°

⇒ ∠ABE + ∠ADC = 180°

⇒ 180° - 2∠AEB + ∠ADC = 180°

⇒ ∠ADC = 180° - 180° + 2∠AEB

⇒ ∠ADC = 2∠AEB .........(3)

Since, ADCF is also a cyclic quadrilateral,

∴ ∠AFC + ∠ADC = 180°

⇒ ∠AFC = 180° - ∠ADC

⇒ ∠AFC = 180° - 2∠AEB [From (3)]

In △ECF,

⇒ ∠EFC + ∠ECF + ∠FEC = 180°

⇒ ∠AFC + ∠ECF + ∠FEC = 180° [From figure, ∠EFC = ∠AFC]

⇒ ∠AFC + ∠ECF + ∠AEB = 180° [∠FEC = ∠AEB (Vertically opposite angles are equal)]

⇒ ∠ECF = 180° - (∠AFC + ∠AEB)

⇒ ∠ECF = 180° - (180° - 2∠AEB + ∠AEB)

⇒ ∠ECF = ∠AEB

⇒ ∠ECF = ∠FEC

∴ EF = FC [As sides opposite to equal angles are equal.]

Hence, proved that EF = FC.

(ii) AF is the angle bisector of ∠A,

∴ ∠BAF = ∠DAF

∴ arc BF = arc DF [As equal arcs subtends equal angles]

∴ BF = DF [Equal arcs have equal chords]

Hence, proved that BF = DF.

Join BC and BD.

In △ABC and △ABD,

⇒ AC = AD [Given]

⇒ AB = AB [Common]

⇒ ∠ACB = ∠ADB [Both = 90°, Angle in semi-circle is a right angle.]

∴ △ ABC ≅ △ ABD [By RHS axiom of congruency]

(i) Since,

△ABC ≅ △ABD

∴ BC = BD [By C.P.C.T.]

∴ Arc BC = Arc BD [Equal chords have equal arcs].

Hence, proved that arc BC = arc BD.

(ii) Since,

△ABC ≅ △ABD

∴ ∠BAC = ∠BAD [By C.P.C.T.]

∴ AB is the bisector of ∠CAD.

Hence, proved that AB is the bisector of ∠CAD.

(a) Given,

arc AC = 2 arc BC

∴ ∠ABC = 2∠BAC

But,

⇒ ∠ABC + ∠BAC = 90° [ΔABC is right angled at C]

⇒ 2∠BAC + ∠BAC = 90°

⇒ 3∠BAC = 90°

⇒ ∠BAC = $\frac{90°}{3}$ = 30°.

Hence, ∠BAC = 30°.

(b) ∠ABC = 2∠BAC = 2 x 30° = 60°.

Hence, ∠ABC = 60°.

In △AEC,

⇒ ∠ACE + ∠CAE + ∠AEC = 180°

⇒ 43° + 62° + ∠AEC = 180°

⇒ ∠AEC = 180° - 105° = 75°.

From figure,

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

⇒ a + ∠AEC = 180° [From figure, ∠AED = ∠AEC]

⇒ a + 75° = 180°

⇒ a = 180° - 75° = 105°.

∠BDC = c [Vertically opposite angles are equal]

∠DBC = 180° - a [Linear pairs]
= 180° - 105°
= 75°.

In △DBC,

⇒ ∠DBC + ∠BCD + ∠BDC = 180° [Angle sum property of triangle]

⇒ 75° + 43° + c = 180°

⇒ 118° + c = 180°

⇒ c = 180° - 118° = 62°.

In △BAF,

⇒ ∠ABF + ∠BAF + ∠AFB = 180° [Angle sum property of triangle]

⇒ a + 62° + b = 180°

⇒ 105° + 62° + b = 180°

⇒ b + 167° = 180°

⇒ b = 180° - 167° = 13°.

Hence, a = 105°, b = 13° and c = 62°.

(i) We know that,

Exterior angle of a cyclic quadrilateral is equal to interior opposite angle.

∠BAD = Exterior ∠BCE = 80°.

From figure,

∠CAD = ∠BAD - ∠BAC = 80° - 25° = 55°.

Hence, ∠CAD = 55°.

(ii) We know that,

Angles in same segment are equal.

∴ ∠CBD = ∠CAD = 55°.

Hence, ∠CBD = 55°.

(iii) We know that,

Angles in same segment are equal.

∴ ∠BDC = ∠BAC = 25°.

AB || DC and BD is transversal.

So, ∠ABD = ∠BDC = 25°. [Alternate angles are equal]

From figure,

∠ABC = ∠ABD + ∠CBD = 25° + 55° = 80°.

In cyclic quadrilateral ABCD,

⇒ ∠ABC + ∠ADC = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ 80° + ∠ADC = 180°

⇒ ∠ADC = 180° - 80° = 100°.

Hence, ∠ADC = 100°.

Since, CP is the bisector of ∠ACB,

∴ ∠ACP = ∠BCP .........(1)

We know that,

Angles in same segment are equal.

∴ ∠ACP = ∠ADP ..........(2)

∠BCP = ∠BDP ...........(3)

From (1) and (2) we get,

∠BCP = ∠ADP ...........(4)

From (3) and (4) we get,

∠ADP = ∠BDP.

∴ DP is the bisector of ∠ADB.

Hence, proved that DP is the bisector of ∠ADB.

We know that,

Angles in same segment are equal.

∴ ∠CAD = ∠CBD = 70°.

From figure,

∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°.

(i) As sum of opposite angles in cyclic quadrilateral = 180°.

In cyclic quadrilateral ABCD,

⇒ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 100° = 180°

⇒ ∠BCD = 180° - 100° = 80°.

Hence, ∠BCD = 80°.

(ii) Since, AD = BC.

∴ ABCD is an isosceles trapezium and AB || DC.

∠DCA = ∠BAC = 30° [Alternate angles]

From figure,

∠BCA = ∠BCD - ∠DCA = 80° - 30° = 50°.

Hence, ∠BCA = 50°.

(iii) As angles in same segment are equal.

∠ABD = ∠DCA = 30°

From figure,

∠ABC = ∠ABD + ∠CBD = 30° + 70° = 100°.

Hence, ∠ABC = 100°.

(iv) As angles in same segment are equal.

∠ADB = ∠BCA = 50°.

Hence, ∠ADB = 50°.

(i) Given,

AD || BC

∴ OD || BC and BD is transversal.

∠ODB = ∠CBD = 32° [Alternate angles are equal]

In △OBD,

OB = OD [Radius of same circle]

∠OBD = ∠ODB = 32°.

Hence, ∠OBD = 32°.

(ii) Given,

AD || BC

∴ AO || BC and OB is transversal.

∠AOB = ∠OBC [Alternate angles are equal]

From figure,

∠OBC = ∠OBD + ∠DBC = 32° + 32° = 64°.

∴ ∠AOB = 64°.

Hence, ∠AOB = 64°.

(iii) In △OAB,

OA = OB [Radius of same circle]

∠OAB = ∠OBA = x (let)

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x + 64° = 180°

⇒ 2x = 180° - 64°

⇒ 2x = 116°

⇒ x = $\frac{116°}{2}$ = 58°

i.e., ∠OAB = 58°.

From figure,

∠DAB = ∠OAB = 58°.

We know that,

Angle in same segment are equal.

∠BED = ∠DAB = 58°.

Hence, ∠BED = 58°.

(i) Given,

∠DAE = 70°

⇒ ∠DAE + ∠BAD = 180° [Linear pairs]

⇒ 70° + ∠BAD = 180°

⇒ ∠BAD = 180° - 70° = 110°.

We know that,

Sum of opposite angles in a cyclic quadrilateral = 180°

⇒ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 110° = 180°

⇒ ∠BCD = 180° - 110° = 70°.

Hence, ∠BCD = 70°.

(ii) We know that,

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

⇒ ∠BOD = 2∠BCD = 2 × 70° = 140°.

Hence, ∠BOD = 140°.

(iii) In △OBD,

OB = OD [Radius of same circle]

∠OBD = ∠ODB = x.

⇒ ∠OBD + ∠ODB + ∠BOD = 180°

⇒ x + x + 140° = 180°

⇒ 2x = 180° - 140°

⇒ 2x = 40°

⇒ x = $\frac{40°}{2}$ = 20°.

∴ ∠OBD = 20°.

Hence, ∠OBD = 20°.