Angle & Cyclic Properties of a Circle — Exercise 18

In the given figure, O is the centre of a circle, ∠OAB = 30° and ∠OCB = 40°. Calculate ∠AOC.

Answer

Join AC.

As, OA = OC = radius of circle.

Let ∠OAC = ∠OCA = x (As 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°.

In the given figure, O is the centre of the circle and ∠AOC = 130°. Find ∠ABC.

Answer

From figure,

⇒ ∠AOC + Reflex ∠AOC = 360°

⇒ 130° + Reflex ∠AOC = 360°

⇒ Reflex ∠AOC = 360° - 130°

⇒ Reflex ∠AOC = 230°.

Arc AC subtends Reflex ∠AOC at center and ∠ABC at another point of circle.

⇒ Reflex ∠AOC = 2 ∠ABC

⇒ 2∠ABC = 230°

⇒ ∠ABC = $\dfrac{230^{\circ}}{2}$

⇒ ∠ABC = 115°.

Hence, the value of ∠ABC = 115°.

In the given figure, O is the centre of the circle and ∠AOB = 110°. Calculate:

(i) ∠ACO

(ii) ∠CAO.

Answer

(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∠ACO

⇒ 110° = 2∠ACO

⇒ ∠ACO = $\dfrac{110^{\circ}}{2}$

⇒ ∠ACO = 55°

Hence, ∠ACO = 55°.

(ii) From figure,

⇒ ∠COA + ∠AOB = 180° [Linear pair]

⇒ ∠COA + 110° = 180°

⇒ ∠COA = 180° - 110°

⇒ ∠COA = 70°

The sum of the three interior angles of any triangle is always 180°.

In ΔAOC,

⇒ ∠COA + ∠ACO + ∠CAO = 180°

⇒ 70° + 55° + ∠CAO = 180°

⇒ 125° + ∠CAO = 180°

⇒ ∠CAO = 180° - 125°

⇒ ∠CAO = 55°.

Hence, ∠CAO = 55°.

In the given figure, AB ∥ DC and ∠BAD = 100°. Calculate :

(i) ∠BCD

(ii) ∠ADC

(iii) ∠ABC.

Answer

(i) We know that:

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

⇒ ∠BAD + ∠BCD = 180°

⇒ ∠BCD = 180° - 100°

⇒ ∠BCD = 80°.

Hence, ∠BCD = 80°.

(ii) Since AB ∥ DC, the angles ∠BAD and ∠ADC are consecutive interior angles along the transversal AD.

Therefore,

⇒ ∠BAD + ∠ADC = 180°

⇒ 100° + ∠ADC = 180°

⇒ ∠ADC = 180° - 100°

⇒ ∠ADC = 80°.

Hence, ∠ADC = 80°.

(iii) We know that:

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

⇒ ∠ABC + ∠ADC = 180°

⇒ ∠ABC + 80° = 180°

⇒ ∠ABC = 180° - 80°

⇒ ∠ABC = 100°.

Hence, ∠ABC = 100°.

In the given figure, ∠ACB = 52° and ∠BDC = 43°. Calculate

(i) ∠ADB

(ii) ∠BAC

(iii) ∠ABC.

Answer

(i) From figure,

∠ADB = ∠ACB = 52° [Angles in the same segment are equal]

∠ADB = 52°.

Hence, ∠ADB = 52°.

(ii) From figure,

∠BAC = ∠BDC = 43° [Angles in the same segment are equal]

∠BAC = 43°.

Hence, ∠BAC = 43°.

(iii) We know that,

The sum of the three interior angles of any triangle is always 180°.

⇒ ∠BAC + ∠ABC + ∠ACB = 180°

⇒ 43° + ∠ABC + 52° = 180°

⇒ ∠ABC + 95° = 180°

⇒ ∠ABC = 180° - 95°

⇒ ∠ABC = 85°.

Hence, ∠ABC = 85°.

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

(i) ∠ABC

(ii) ∠BCO

(iii) ∠OAB

(iv) ∠BCA

Answer

(i) From figure,

OA = OC [Radius of same circle]

In Δ AOC,

∠OCA = ∠OAC = 50° (As angles opposite to equal sides in a triangle are equal)

We know that,

The sum of the three interior angles of any triangle is always 180°.

∠AOC + ∠OCA + ∠OAC = 180°

∠AOC + 50° + 50° = 180°

∠AOC = 180° − 50° − 50° = 80°.

We know that,

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

∠ABC = $\dfrac{1}{2}$ ∠AOC

∠ABC = $\dfrac{1}{2}$ 80°

∠ABC = 40°.

Hence, ∠ABC = 40°.

(ii) From figure,

∠BOC = ∠AOB − ∠AOC = 140° − 80° = 60°.

OB = OC (Radius of same circle)

Let ∠OBC = ∠OCB = x (As angles opposite to equal sides are equal)

We know that,

The sum of the three interior angles of any triangle is always 180°.

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ x + x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120^{\circ}}{2}$

⇒ x = 60°.

Hence, ∠BCO = 60°.

(iii) In ∆AOB, we have

OA = OB (radius of same 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 = $\dfrac{40^{\circ}}{2}$

⇒ ∠OAB = 20°.

Hence, ∠OAB = 20°.

(iv) We know that,

⇒ ∠BCO = 60°

⇒ ∠OCA = 50°

From figure,

⇒ ∠BCA = ∠BCO + ∠OCA

⇒ ∠BCA = 60° + 50°

⇒ ∠BCA = 110°.

Hence, ∠BCA = 110°.

In the given figure, ∠BAD = 70°, ∠ABD = 50° and ∠ADC = 80°. Calculate :

(i) ∠BDC

(ii) ∠BCD

(iii) ∠BCA.

Answer

(i) By angle sum property of a triangle we get,

In ΔABD,

⇒ ∠BAD + ∠ABD + ∠ADB = 180°

⇒ 70° + 50° + ∠ADB = 180°

⇒ 120° + ∠ADB = 180°

⇒ ∠ADB = 180° - 120°

⇒ ∠ADB = 60°.

From figure,

⇒ ∠BDC = ∠ADC - ∠ADB

⇒ ∠BDC = 80° - 60°

⇒ ∠BDC = 20°.

Hence, ∠BDC = 20°.

(ii) We know that,

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

⇒ ∠BAD + ∠BCD = 180°

⇒ 70° + ∠BCD = 180°

⇒ ∠BCD = 180° - 70°

⇒ ∠BCD = 110°.

Hence, ∠BCD = 110°.

(iii) We know that,

Angles in the same segment of a circle are equal.

⇒ ∠BCA = ∠ADB

⇒ ∠BCA = 60°.

Hence, ∠BCA = 60°.

In the given figure, O is the centre of the circle. If ∠ADC = 140°, find ∠BAC.

Answer

From figure,

∠ACB = 90° [Angle in a semicircle is a right angle]

We know that,

In a cyclic quadrilateral, the sum of opposite angles is 180°.

⇒ ∠ABC + ∠ADC = 180°

⇒ ∠ABC + 140° = 180°

⇒ ∠ABC = 180° - 140°

⇒ ∠ABC = 40°.

By angle sum property of a triangle we get,

⇒ ∠BAC + ∠ABC + ∠ACB = 180°

⇒ ∠BAC + 40° + 90° = 180°

⇒ ∠BAC + 130° = 180°

⇒ ∠BAC = 180° - 130°

⇒ ∠BAC = 50°.

Hence, ∠BAC = 50°.

PQRS is a cyclic quadrilateral. Given ∠QPS = 73°, ∠PQS = 55° and ∠PSR = 82°, calculate ∠QRS, ∠RQS and ∠PRQ.

Answer

We know that,

In a cyclic quadrilateral, the sum of opposite angles is 180°.

⇒ ∠QPS + ∠QRS = 180°

⇒ 73° + ∠QRS = 180°

⇒ ∠QRS = 180° - 73°

⇒ ∠QRS = 107°.

From figure,

⇒ ∠PSR + ∠PQR = 180°

⇒ ∠PSR + ∠PQS + ∠RQS = 180°

⇒ 82° + 55° + ∠RQS = 180°

⇒ 137° + ∠RQS = 180°

⇒ ∠RQS = 180° - 137°

⇒ ∠RQS = 43°.

By angle sum property of a triangle we get,

⇒ ∠PSQ + ∠PQS + ∠QPS = 180°

⇒ ∠PSQ + 55° + 73° = 180°

⇒ ∠PSQ + 128° = 180°

⇒ ∠PSQ = 180° - 128°

⇒ ∠PSQ = 52°.

∠PSQ = ∠PRQ = 52° [Angles in the same segment]

Hence, ∠QRS = 107°, ∠RQS = 43° and ∠PRQ = 52°.

In the given figure, O is the centre of the circle and ΔABC is equilateral. Find :

(i) ∠BDC

(ii) ∠BEC.

Answer

(i) Since ABC is an equilateral triangle so,

∠A = ∠B = ∠C = 60°.

From figure,

∠BDC = ∠BAC (Angles in same segment are equal.)

∴ ∠BDC = 60°.

Hence, ∠BDC = 60°.

(ii) BDCE is a cyclic quadrilateral. Hence, sum of the opposite angles = 180°.

⇒ ∠BDC + ∠BEC = 180°

⇒ 60° + ∠BEC = 180°

⇒ ∠BEC = 180° - 60° = 120°.

Hence, ∠BEC = 120°.

In the given figure, O is the centre of the circle and ∠AOC = 160°. Prove that 3∠y − 2∠x = 140°.

Answer

We know that,

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

∠AOC = 2∠ABC

∠AOC = 2x

∠x = $\dfrac{1}{2}$ ∠AOC = $\dfrac{160^{\circ}}{2}$ = 80°.

In a cyclic quadrilateral, the sum of opposite angles is 180°.

In quadrilateral ABCD,

⇒ ∠ABC + ∠ADC = 180°

⇒ ∠x + ∠y = 180°

⇒ 80° + ∠y = 180°

⇒ ∠y = 100°.

Substitute values in L.H.S of equation 3∠y − 2∠x = 140° :

= 3(100°) − 2(80°)

= 300° − 160°

= 140°.

As, L.H.S = R.H.S

Hence, proved that 3∠y − 2∠x = 140°.

In the given figure, O is the centre of the circle. If ∠CBD = 25° and ∠APB = 120°, find ∠ADB.

Answer

From figure,

⇒ ∠CPB + ∠APB = 180° [Linear pair]

⇒ ∠CPB = 180° - ∠APB

⇒ ∠CPB = (180° − 120°) = 60°.

By angle sum property of a triangle we get,

⇒ ∠PCB + ∠CPB + ∠PBC = 180°

⇒ ∠PCB + 60° + 25° = 180°

⇒ ∠PCB + 85° = 180°

⇒ ∠PCB = 180° - 85°

⇒ ∠PCB = 95°

⇒ ∠ADB = ∠ACB = 95° [Angles in same segment of a circle are equal]

Hence, the value of ∠ADB = 95°.

In the given figure, O is the centre of the circle, ∠BAD = 75° and chord BC = chord CD. Find : (i) ∠BOC (ii) ∠OBD (iii) ∠BCD

Answer

(i) As equal chords of a circle subtend equal angles at the center and chord BC = chord CD, so ∠BOC = ∠COD.

∠BOD = 2 × ∠BAD

∠BOD = 2 × 75°

∠BOD = 150°

∠BOC = $\dfrac{1}{2}$ ∠BOD

= $\dfrac{1}{2}$ × 150° = 75°.

Hence, the value of ∠BOC = 75°.

(ii) Join BD.

Since, OB = OD

∴ ∠OBD = ∠ODB = x

Since sum of angles of triangle = 180°

In △OBD

⇒ ∠BOD + ∠OBD + ∠ODB = 180°

⇒ 150° + x + x = 180°

⇒ 150° + 2x = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = 15°.

Hence, the value of ∠OBD = 15°.

(iii) ABCD is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.

We know that sum of opposite angles of a cyclic quadrilateral = 180°.

⇒ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 75° = 180°

⇒ ∠BCD = 180° - 75°

⇒ ∠BCD = 105°.

Hence, the value of ∠BCD = 105°.

In the figure given, O is the centre of the circle. ∠DAE = 70°, find giving suitable reasons, the measure of :

(i) ∠BCD

(ii) ∠BOD

(iii) ∠OBD

Answer

(i) Given,

∠DAE = 70°

From figure,

⇒ ∠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 = $\dfrac{40^{\circ}}{2}$ = 20°.

∴ ∠OBD = 20°.

Hence, ∠OBD = 20°.

In the given figure, AOB is a diameter of the circle with centre O and ∠AOC = 100°, find ∠BDC.

Answer

From figure,

∠AOC + ∠BOC = 180° [Linear pair]

100° + ∠BOC = 180°

∠BOC = 180° - 100°

∠BOC = 80°.

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.

⇒ ∠BOC = 2∠BDC

⇒ 80° = 2∠BDC

⇒ ∠BDC = $\dfrac{80^{\circ}}{2}$ = 40°

Hence, ∠BDC = 40°.

In the given figure, find whether the points A, B, C, D are concyclic when:

(i) x = 70

(ii) x = 80

Answer

From the figure,

∠ABC + ∠CBE = 180° [Linear pair]

∠ABC + 110° = 180°

∠ABC = 70°

For the points A, B, C, D to be concyclic, ∠ABC + ∠ADC must be equal to 180°.

(i) When x = 70

x° + ∠ADC = 180° [Linear pair]

∠ADC = 180° − x° = 180° - 70° = 110°.

∠ABC + ∠ADC = 70° + 110° = 180°

Since the sum of opposite angles is 180°, the points A, B, C, D are concyclic when x = 70.

Hence, yes the points A, B, C, D are concyclic when x = 70.

(ii) When x = 80

x° + ∠ADC = 180° [Linear pair]

∠ADC = 180° − 80° = 100°

Check for con-cyclicity,

∠ABC + ∠ADC = 70° + 100° = 170°

Since the sum of opposite angles is not 180°, the points A, B, C, D are not concyclic when x = 80.

Hence, no the points A, B, C, D are not concyclic when x = 80.

In the adjoining figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find: (i) ∠BCD (ii) ∠ADB Hence, show that AC is a diameter.

Answer

(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°.

From figure,

∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°.

Since, angle in a semi-circle is a right angle. Thus, AC is the diameter.

Hence, ∠ADB = 45° and proved that AC is a diameter.

In the given figure, ABCD is a cyclic quadrilateral in which ∠CAD = 25°, ∠ADB = 35° and ∠ABD = 50°. Calculate:

(i) ∠CBD

(ii) ∠CAB

(iii) ∠ACB

Answer

(i) We know that,

Angles in same segment are equal.

∠CBD = ∠CAD = 25°

Hence, ∠CBD = 25°.

(ii) We know that,

Angles in same segment are equal.Therefore,

⇒ ∠ACD = ∠ABD = 50°

⇒ ∠ACB = ∠ADB = 35°

From figure

⇒ ∠BCD = ∠ACD + ∠ACB

⇒ ∠BCD = 35° + 50°

⇒ ∠BCD = 85°.

We know that,

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

⇒ ∠DAB + ∠DCB = 180°

⇒ ∠DAB = 180° - 85°

⇒ ∠DAB = 95°.

From figure,

⇒ ∠CAB = ∠DAB - ∠DAC

= 95° - 25°

= 70°.

Hence, ∠CAB = 70°.

(iii) Angles in same segment are equal. Therefore,

∠ACB = ∠ADB = 35°

Hence, ∠ACB = 35°.

In the figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find :

(i) ∠CAD

(ii) ∠CBD

(iii) ∠ADC

Answer

(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) Since, AB ∥ DC,

∠ACD = ∠BAC = 25° [Alternate angles are equal]

In triangle ADC,

By angle sum property of triangle,

⇒ ∠ADC + ∠CAD + ∠ACD = 180°

⇒ ∠ADC + 55° + 25° = 180°

⇒ ∠ADC + 80° = 180°

⇒ ∠ADC = 180° - 80°

⇒ ∠ADC = 100°.

Hence, ∠ADC = 100°.

In the adjoining figure of a circle with centre O and diameter AD, ∠BED = 70° and BC is parallel to AD. Find:

(i) ∠BAD

(ii) ∠BOD

(iii) ∠DBC

(iv) ∠DCF

Answer

(i) Given,

∠BED = 70°

We know that,

Angles in the same segment of a circle are equal.

∠BAD = ∠BED = 70°.

Hence, ∠BAD = 70°.

(ii) We know that,

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

Therefore,

∠BOD = 2∠BAD = 2 × 70° = 140°.

Hence, ∠BOD = 140°.

(iii) We know that,

Angle in a semi-circle is a right angle triangle.

∠ABD = 90°

In △ABD,

⇒ ∠BAD + ∠BDA + ∠ABD = 180°

⇒ 70° + ∠BDA + 90° = 180°

⇒ 160° + ∠BDA = 180°

⇒ ∠BDA = 180° - 160°

⇒ ∠BDA = 20°

From figure,

BC || AD

∴ ∠DBC = ∠BDA = 20° (Alternate angles are equal).

Hence, ∠DBC = 20°.

(iv) The figure ABCD is a cyclic quadrilateral

We know that,

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

∠DCF = ∠BAD = 70°.

Hence, ∠DCF = 70°.

In the given figure, AB is a diameter of a circle with centre O and chord ED is parallel to AB and ∠EAB = 65°. Calculate : (i) ∠EBA (ii) ∠BED (iii) ∠BCD

Answer

(i) We know that,

Angle in semi-circle is a right angle.

∴ ∠AEB = 90°.

In △AEB,

⇒ ∠AEB + ∠EBA + ∠EAB = 180°

⇒ 90° + ∠EBA + 65° = 180°

⇒ 155° + ∠EBA = 180°

⇒ ∠EBA = 180° - 155° = 25°.

Hence, ∠EBA = 25°.

(ii) ∠BED = ∠EBA = 25° Alternate interior angles ED ∥ AB, EB as a transversal

Hence, ∠BED = 25°.

(iii) As, AB ∥ ED

∴ ∠DEB = ∠EBA = 25° [Alternate angles]

BCDE is a cyclic quadrilateral.

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

⇒ 25° + ∠BCD = 180°

⇒ ∠BCD = 180° - 25° = 155°.

Hence, ∠BCD = 155°.

In the given figure, O is the centre of a circle and ABE is a straight line. If ∠CBE = 55°, find :

(i) ∠ADC

(ii) ∠ABC

(iii) the value of x.

Answer

(i) We know that,

In a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.

∠ADC = ∠CBE = 55°

Hence, ∠ADC = 55°.

(ii) From figure,

∠CBE + ∠ABC = 180° [Linear pairs]

∠ABC = 180° - 55°

∠ABC = 125°.

Hence, ∠ABC = 125°.

(iii) 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.

∠AOC = 2∠ADC

∠AOC = 2(55°)

∠AOC = 110°

The reflex angle AOC,

x° = 360° - angle AOC

x° = 360° - 110°

x° = 250°.

Hence, x = 250.

In the given figure AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E, prove that Δ AEB is isosceles.

Answer

We know that,

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

∠EDC = ∠A ....(i)

∠DCE = ∠B ....(ii)

AB ∥ CD

∠EDC = ∠B [Corresponding angles] ......(iii)

∠DCE = ∠A [Corresponding angles] .......(iv)

From (i), (ii), (iii) and (iv) we get :

∴ ∠A = ∠B

BE = AE [Sides opposite to equal angles are equal]

Hence, proved that ΔAEB is isosceles.

In the given figure, the two circles intersect at P and Q. If ∠A = 80° and ∠D = 84°, calculate :

(i) ∠QBC

(ii) ∠BCP

Answer

PQAD is a cyclic quadrilateral as all vertices lie on the circumference of the circle.

Sum of opposite angles of cyclic quadrilateral = 180°

⇒ ∠DAQ + ∠DPQ = 180°

⇒ 80° + ∠DPQ = 180°

⇒ ∠DPQ = 180° - 80°

⇒ ∠DPQ = 100°.

Also,

⇒ ∠PDA + ∠PQA = 180°

⇒ 84° + ∠PQA = 180°

⇒ ∠PQA = 180° - 84°

⇒ ∠PQA = 96°.

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

(i) ∠QBC = ∠DPQ = 100°.

Hence, ∠QBC = 100°.

(ii) ∠BCP = ∠PQA = 96°.

Hence, ∠BCP = 96°.

In the given figure, O is the centre of the circle. If ∠AOD = 140° and ∠CAB = 50°, calculate :

(i) ∠EDB

(ii) ∠EBD

Answer

(i) We know that,

An Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. ∠EDB is the exterior angle at vertex D. The interior opposite angle is ∠CAB.

⇒ ∠EBD = ∠CAB = 50°

Hence, the value of ∠EDB = 50°.

(ii) We know that,

⇒ OD = OB [radii of same circle]

From figure,

⇒ ∠AOD + ∠DOB = 180° [Linear pairs]

⇒ 140° + ∠DOB = 180°

⇒ ∠DOB = 180° - 140°

⇒ ∠DOB = 40°

In ΔODB,

By angle sum property of triangle,

⇒ ∠DOB + ∠ODB + ∠OBD = 180°

⇒ 40° + 2∠OBD = 180°

⇒ 2∠OBD = 180° - 40°

⇒ 2∠OBD = 140°

⇒ ∠OBD = $\dfrac{180^{\circ}}{2}$

⇒ ∠OBD = 70°

From figure,

⇒ ∠OBD + ∠EBD = 180° [Linear pairs]

⇒ 70° + ∠EBD = 180°

⇒ ∠EBD = 180° - 70°

⇒ ∠EBD = 110°

Hence, the value of ∠EBD = 110°.

In the given figure, AB is a diameter of a circle with centre O. If ADF and CBF are straight lines, meeting at F such that ∠BAD = 35° and ∠BFD = 25°, find :

(i) ∠DCB

(ii) ∠DBC

(iii) ∠BDC

Answer

(i) We know that,

Angles in same segment are equal.

∠DCB = ∠BAD = 35°

Hence, the value of ∠DCB = 35°.

(ii) From figure,

∠ADB = 90° (∵ Angle in semicircle is 90°.)

∴ ∠BDF = 90°

In △DBF,

⇒ ∠DBF + ∠DFB + ∠BDF = 180° (∵ By angle sum triangle property).

⇒ ∠DBF = 180° - (∠DFB + ∠BDF)

⇒ ∠DBF = 180° - (90° + 25°)

⇒ ∠DBF = 65°.

⇒ ∠DBF + ∠DBC = 180° (Linear pairs)

⇒ ∠DBC = 180° - 65°

⇒ ∠DBC = 115°.

Hence, the value of ∠DBC = 115°.

(iii) In △DCB,

⇒ ∠DCB + ∠CBD + ∠BDC = 180° (∵ By angle sum triangle property).

⇒ ∠BDC = 180° - (∠DCB + ∠CBD)

⇒ ∠BDC = 180° - (35° + 115°)

⇒ ∠BDC = 180° - (35° + 115°)

⇒ ∠BDC = 30°.

Hence, the value of ∠BDC = 30°.

In the given figure, the straight lines AB and CD pass through the centre O of the circle. If ∠AOD = 75° and ∠OCE = 40°, find :

(i) ∠CDE

(ii) ∠OBE.

Answer

(i) Since, AB and CD pass through the center, thus they are the diameters of circle.

In △CED,

∠CED = 90° (∵ angle in semicircle is 90°.)

We know that sum of angles of a triangle is 180°.

⇒ ∠CED + ∠DCE + ∠CDE = 180°.

⇒ 90° + 40° + ∠CDE = 180°

⇒ ∠CDE + 130° = 180°

⇒ ∠CDE = 180° - 130°

⇒ ∠CDE = 50°.

Hence, ∠CDE = 50°.

(ii) From figure,

⇒ ∠AOD + ∠DOB = 180° (Linear pairs)

⇒ 75° + ∠DOB = 180°

⇒ ∠DOB = 180° - 75°

⇒ ∠DOB = 105°.

In △DOB,

∠ODB = ∠CDE = 50°

We know that,

Sum of angles of a triangle is 180°.

⇒ ∠DOB + ∠ODB + ∠DBO = 180°.

⇒ 105° + 50° + ∠DBO = 180°

⇒ ∠DBO + 155° = 180°

⇒ ∠DBO = 180° - 155°

⇒ ∠DBO = 25°.

From figure,

∠OBE = ∠DBO

∴ ∠OBE = 25°.

Hence, ∠OBE = 25°.

In the adjoining figure, AB = AC = CD and ∠ADC = 35°. Calculate :

(i) ∠ABC

(ii) ∠BEC

Answer

(i) Given,

AC = CD

In triangle ACD,

∠DAC = ∠ADC = 35° [Angles opposite to equal sides in a triangle are equal]

In △ACD,

⇒ ∠DAC + ∠ADC + ∠ACD = 180°

⇒ 35° + 35° + ∠ACD = 180°

⇒ 70° + ∠ACD = 180°

⇒ ∠ACD = 180° - 70° = 110°.

From figure,

⇒ ∠ACB + ∠ACD = 180° [Linear pair]

⇒ ∠ACB + 110° = 180°

⇒ ∠ACB = 180° - 110° = 70°.

Given,

AB = AC

∴ ∠ABC = ∠ACB = 70°. [As angles opposite to equal sides are equal]

Hence, ∠ABC = 70°.

(ii) In △ABC,

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

⇒ ∠BAC + 70° + 70° = 180°

⇒ ∠BAC + 140° = 180°

⇒ ∠BAC = 180° - 140° = 40°.

We know that,

Angles in same segment are equal.

⇒ ∠BEC = ∠BAC = 40°.

Hence, ∠BEC = 40°.

The exterior angles B and C in ΔABC are bisected to meet at a point P. Prove that ∠BPC = 90° − $\dfrac{A}{2}$. Is ABPC a cyclic quadrilateral ?

Answer

We know that,

An exterior angle of a triangle equals to the sum of the two interior opposite angles.

∠CBD = ∠A + ∠C

∠BCE = ∠A + ∠B.

Since BP bisects the exterior angle ∠CBD :

∠PBC = $\dfrac{1}{2}$ ∠CBD

Similarly, since CP bisects the exterior angle ∠BCE :

∠PCB = $\dfrac{1}{2}$ ∠BCE

In △BPC,

By angle sum property of triangle,

$\Rightarrow ∠PBC + ∠PCB + ∠BPC = 180^{\circ} \\[1em] \Rightarrow \dfrac{1}{2}∠CBD + \dfrac{1}{2}∠BCE + ∠BPC = 180^{\circ} \\[1em] \Rightarrow \dfrac{1}{2}(∠A + ∠C) + \dfrac{1}{2}(∠A + ∠B) + ∠BPC = 180^{\circ} \\[1em] \Rightarrow ∠A + \dfrac{1}{2}(∠C + ∠B) + ∠BPC = 180^{\circ} \\[1em] \Rightarrow \dfrac{1}{2}(∠B + ∠C) + ∠BPC = 180^{\circ} - A \\[1em] \Rightarrow \dfrac{1}{2}(180^{\circ} - A) + ∠BPC = 180^{\circ} - A \\[1em] \Rightarrow 90^{\circ} - \dfrac{A}{2} + ∠BPC = 180^{\circ} - A \\[1em] \Rightarrow ∠BPC = 180^{\circ} - A - \Big[90^{\circ} - \dfrac{A}{2}\Big] \\[1em] \Rightarrow ∠BPC = 180^{\circ} - A - 90^{\circ} + \dfrac{A}{2}\\[1em] \Rightarrow ∠BPC = 90^{\circ} - \dfrac{A}{2}.$

For A, B, P, C to be concyclic we would need ∠A + ∠BPC = 180° But:

∵ ∠A + ∠BPC = ∠A + $90^{\circ} - \dfrac{∠A}{2} = 90^{\circ} + \dfrac{∠A}{2}$

Hence, proved that $∠BPC = 90° - \dfrac{∠A}{2}$ and ABPC is not cyclic.

In the given figure, O is the centre of the circle. If QR = OP and ∠ORP = 20°, find the value of x giving reasons.

Answer

Given,

⇒ QR = OP

⇒ QR = OQ (As, OP = OQ = Radius of circle)

⇒ ∠QOR = ∠ORQ = 20° (Angles opposite to equal sides of a triangle are equal)

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

∴ ∠OQP = ∠QOR + ∠ORQ = 20° + 20° = 40°.

As OP = OQ, ∠OPQ = ∠OQP

⇒ ∠OPQ = 40°

⇒ ∠OPR = 40°.

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

In triangle OPR,

∴ x° = ∠OPR + ∠ORP = 40° + 20° = 60°.

Hence, the value of x = 60.

In the given figure, I is the incentre of Δ ABC. Here AI produced meets the circumcircle of Δ ABC at D. If ∠ABC = 55° and ∠ACB = 65°, calculate :

(i) ∠BCD

(ii) ∠CBD

(iii) ∠DCI

(iv) ∠BIC

Answer

(i) Join BI and CI.

In △ABC,

⇒ ∠BAC + ∠ABC + ∠ACB = 180° (∵ sum of angles = 180°.)

⇒ ∠BAC + 55° + 65° = 180°

⇒ ∠BAC + 120° = 180°

⇒ ∠BAC = 180° - 120°

⇒ ∠BAC = 60°.

I is the incentre,

∴ I lies on the bisectors of angle of the △ABC,

∴ ∠BAD = ∠CAD = $\dfrac{60^{\circ}}{2}$ = 30°.

∠BCD = ∠BAD = 30°. (∵ angles in same segment are equal.)

Hence, the value of ∠BCD = 30°.

(ii) Similarly,

∠CBD = ∠CAD = 30°. (∵ angles in same segment are equal.)

Hence, the value of ∠CBD = 30°.

(iii) The line CI bisects ∠C (∵ I lies on the bisectors of angle of the △ABC).

∴ ∠BCI = $\dfrac{65^{\circ}}{2} = 32 \dfrac{1}{2}^{\circ}$.

From figure,

∠DCI = ∠BCD + ∠BCI = 30° + $32 \dfrac{1}{2}^{\circ}$

= 62.5°.

Hence, the value of ∠DCI = 62.5°.

(iv) ∠IBC = $\dfrac{55}{2}^{\circ}$ = 27.5°

∠ICB = $\dfrac{65}{2}^{\circ}$ = 32.5°

In triangle BIC,

By angle sum property of triangle,

∠BIC + ∠IBC + ∠ICB = 180°

∠BIC = 180° - (∠IBC + ∠ICB)

= 180° - (27.5° + 32.5°)

= 180° - 60°

= 120°.

Hence, the value of ∠BIC = 120°.

In the given figure, ∠DBC = 58° and BD is a diameter of the circle. Calculate :

(i) ∠BDC

(ii) ∠BEC

(iii) ∠BAC

Answer

(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°.

In the given figure, ABCDE is a pentagon inscribed in a circle such that AC is a diameter and side BC ∥ AE. If ∠BAC = 50°, find giving reasons :

(i) ∠ACB

(ii) ∠EDC

(iii) ∠BEC

Hence prove that BE is also a diameter.

Answer

(i) ∠ABC = 90° [Angle in semicircle]

In ∆ABC,

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

⇒ 90° + 50° + ∠ACB = 180°

⇒ 140° + ∠ACB = 180°

⇒ ∠ACB = 40°

Hence, ∠ACB = 40°.

(ii) ∠EAC = ∠ACB = 40° [Alternate angles, AC transversal to parallel lines AE and BC]

∠EAC + ∠EDC = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

40° + ∠EDC = 180°

∠EDC = 140°

Hence, ∠EDC = 140°.

(iii) ∠EBC + ∠EDC = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

140° + ∠EBC = 180°

∠EBC = 40°

In ∆EBC,

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

⇒ ∠BEC + 90° + 40° = 180°

⇒ ∠BEC = 180° - 130°

⇒ ∠BEC = 50°

In ∆EAB,

⇒ ∠EAB = ∠EAC + ∠BAC

= 40° + 50° = 90°

We know that, if an angle of a triangle in a circle is 90° then, the hypotenuse must be the diameter of the circle.

Hence, ∠BEC = 50° and BE is the diameter of the circle.

In the given figure, O is the centre of the circle and AB is a diameter. If AC = BD and ∠AOC = 72°, find:

(i) ∠ABC

(ii) ∠BAD

(iii) ∠ABD

Answer

(i) 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.

∠ABC = $\dfrac{1}{2}$ ∠AOC

∠ABC = $\dfrac{72^{\circ}}{2}$ = 36°

Hence, ∠ABC = 36°.

(ii) Given,

AC = BD

∠BAC = ∠ABC = 36° [Angles opposite to equal sides are equal]

Hence, ∠BAC = 36°.

(iii) In ∆ABD,

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

∠ABD + 36° + 90° = 180°

∠ABD + 126° = 180°

∠ABD = 180° - 126°

∠ABD = 54°

Hence, ∠ABD = 54°.

In the given figure, A, C, B, D are points on the circle with centre O. Given ∠ABC = 62°. Find :

(i) ∠ADC

(ii) ∠CAB

Answer

(i) From figure,

⇒ ∠ADC = ∠ABC (Angles in same segment are equal)

⇒ ∠ADC = 62°.

Hence, ∠ADC = 62°.

(ii) We know that,

Angle in a semi-circle is a right angle.

∠ACB = 90°

Using angle sum property,

⇒ ∠CAB + ∠ACB + ∠ABC = 180°

⇒ ∠CAB + 90° + 62° = 180°

⇒ ∠CAB + 152° = 180°

⇒ ∠CAB = 180° - 152°

⇒ ∠CAB = 28°.

Hence, ∠CAB = 28°.