myhelper: RD Sharma 2020 solution class 9 chapter 15 Circles FBQS

RD Sharma 2020 solution class 9 chapter 15 Circles FBQS

FBQS

Page-15.110

Question 1:

Fill In The Blanks

AD is a diameter, ,of circle and AB is chord. If AD = 34 cm, AB = 30 cm, then BD = ____________

Answer 1:

Given:
AD is a diameter
AB is chord
AD = 34 cm
AB = 30 cm

Let O be the centre of the circle.
AO = OD = 17 cm ...(1)

Let OL is a line perpendicular to AB, where L is the point on AB.
Then, AL = LB = 15 cm ...(2) (∵ a perpendicular from the centre of the circle to the chord, bisects the chord)

In ∆ALO,
Using pythagoras theorem,
AL² + LO² = AO²
⇒ 15² + LO² = 17² (From (1) and (2))
⇒ 225 + LO² = 289
⇒ LO² = 289 − 225
⇒ LO² = 64
⇒ LO = 8 cm

Now, In ∆ABD,
Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.
Therefore, LO = $\frac{1}{2}$ BD
⇒ BD = 2LO
⇒ BD = 2(8)
⇒ BD = 16 cm

Hence, BD = 16 cm.

Question 2:

Fill In The Blanks

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30cm, then the distance of AB from the centre of the circle is ________.

Answer 2:

Given:
AD is a diameter
AB is chord
AD = 34 cm
AB = 30 cm

Let O be the centre of the circle.
AO = OD = 17 cm ...(1)

Let OL is a line perpendicular to AB, where L is the point on AB.
Then, AL = LB = 15 cm ...(2) (∵ a perpendicular from the centre of the circle to the chord, bisects the chord)

In ∆ALO,
Using pythagoras theorem,
AL² + LO² = AO²
⇒ 15² + LO² = 17² (From (1) and (2))
⇒ 225 + LO² = 289
⇒ LO² = 289 − 225
⇒ LO² = 64
⇒ LO = 8 cm

Thus, the distance of the chord from the centre is 8 cm.

Hence, the distance of AB from the centre of the circle is 8 cm.

Question 3:

Fill In The Blanks

If AB = 12 cm, BC = 16 cm, and AB is perpendicular to BC, then the radius of the circle passing through the points A,B and C is _________.

Answer 3:

Given:
AB = 12 cm
BC = 16 cm
AB is perpendicular to BC

Since, AB is perpendicular to BC
Therefore, the circle formed by joining A, B and C is a circle with diameter AC.

In ∆ABC,
Using pythagoras theorem,
AB² + BC² = AC²
⇒ 12² + 16² = AC² (given)
⇒ 144 + 256 = AC²
⇒ AC²= 400
⇒ AC = 20

Thus, the diameter of the circle is 20 cm.
Therefore, the radius of the circle is half of the diameter of the circle.
Radius = $\frac{1}{2} (20)$ =10 cm

Hence, the radius of the circle passing through the points A, B and C is 10 cm.

Question 4:

Fill In The Blanks

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscibing it and ∠ADC = 140^∘,then ∠ BAC = ________.

Answer 4:

Given:
ABCD is a cyclic quadrilateral
AB is a diameter of the circle circumscribing ABCD
∠ADC = 140^∘

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

Thus, ∠ADC + ∠ABC = 180°
⇒ 140° + ∠ABC = 180°
⇒ ∠ABC = 180° − 140°
⇒ ∠ABC = 40° ...(1)

Since AB is the diameter of the circle
Therefore, ∠ACB = 90° (angle in the semi circle) ...(2)

In ∆ABC,
∠BAC + ∠ACB + ∠ABC = 180° (angle sum property)
⇒ ∠BAC + 90° + 40° = 180° (From (1) and (2))
⇒ ∠BAC + 130° = 180°
⇒ ∠BAC = 180° − 130°
⇒ ∠BAC = 50°

Hence, ∠BAC = 50°.

Question 5:

Fill In The Blanks

Two chords AB and CD of a circle are each at a distance of 6 cm from the centre. the ratio of their lengths is ________.

Answer 5:

Given:
Two chords AB and CD of a circle are each at a distance of 6 cm from the centre.

The chords which are equidistant from the centre of the circle are of equal length.

Hence, the ratio of their length is 1 : 1.

Page-15.111

Question 6:

Fill In The Blanks

If two equals chords AB and AC of a circle with centre O are on the opposite sides of OA, then ∠OAB = ____________ .

Answer 6:

Given:
AB =AC
AB and AC lies on the opposite sides of OA

The chords of equal length, are equidistant from the centre of the circle.

In ∆OAB and ∆OAC,
AB = AC (given)
OA = OA (common)
OB = OC (radius of the circle)

By SSS property,
∆OAB ≅ ∆OAC

Therefore, ∠OAB = ∠OAC (by C.P.C.T.)

Hence, ∠OAB = ∠OAC.

Question 7:

Fill In The Blanks

Two congruent circle with centres O and O' intesect at two points P and Q. then then, ∠POQ :∠PO'Q = ____________.

Answer 7:

Given:
Two congruent circle with centres O and O' intersect at two points P and Q.

In ∆OPQ and ∆O'PQ,
OP = O'P (radius)
PQ = PQ (common)
OQ = O'Q (radius)

By SSS property,
∆OPQ ≅ ∆O'PQ

Therefore, ∠POQ = ∠PO'Q (by C.P.C.T.)

Hence, ∠POQ : ∠PO'Q = 1 : 1.

Question 8:

Fill In The Blanks

If AOB is a diameter of a circle and C is a point on the circle, then the AC² + BC²= ____________.

Answer 8:

Given:
AOB is a diameter of a circle
C is a point on the circle

Since, AOB is the diameter of the circle
Therefore, ∠ACB = 90° (angle in the semi circle) ...(2)

In right angled ∆ABC,
Using pythagoras theorem.
AC² + BC² = AB²

Hence, AC² + BC²= AB².

Question 9:

Fill In The Blanks

If O is the circumcentre of ΔABC and D is the mid-point of the base BC, then ∠BOD = _______________.

Answer 9:

Given:
O is the circumcentre of ∆ABC
D is the mid-point of the base BC

In ∆BOD and ∆COD,
OB = OC (radius)
BD = CD (D is the mid-point of the base BC)
OD = OD (common)

By SSS property,
∆BOD ≅ ∆COD

Therefore, ∠BOD = ∠COD (by C.P.C.T.)
⇒ ∠BOC = ∠BOD + ∠COD
⇒ ∠BOC = 2∠BOD ..(1)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠BOC = 2∠BAC
⇒ 2∠BOD = 2∠BAC
⇒ ∠BOD = ∠BAC

Hence, ∠BOD = ∠BAC.

Question 10:

Fill In The Blanks

If O is the circumcentre of ΔABC, then ∠OBC + ∠BAC = __________.

Answer 10:

Given:
O is the circumcentre of ∆ABC

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠BOC = 2∠BAC ...(1)

In ∆OBC ,
OB = OC (radius)
Thus, ∠OBC = ∠OCB ...(2)

Also, ∠OBC + ∠OCB + ∠BOC = 180° (angle sum property)
⇒ 2∠OBC + 2∠BAC = 180° (from (1) and (2))
⇒ ∠OBC + ∠BAC = 90°

Hence, ∠OBC + ∠BAC = 90°.

Question 11:

Fill In The Blanks

A chord of a circle is equal to its radius. The angle subtended by this chord at a point in major segment is ___________.

Answer 11:

Given:
A chord of a circle is equal to its radius

Let AB is a chord and O is the centre of the circle.

AB = OA = OB (∵ Chord is equal to the radius)
⇒ ∆ABO is equilateral triangle

Thus, ∠AOB = 60° ...(1)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOB = 2∠ADB , where D is any point on the major segment of the circle
⇒ 2∠ADB = 60° (from (1) and (2))
⇒ ∠ADB = 30°

Hence, the angle subtended by the chord at a point in major segment is 30°.

Question 12:

Fill In The Blanks

If a pair of opposite sides of a quadrilateral are equal, then its diagonals are ___________..

Answer 12:

Given:
A pair of opposite sides of a quadrilateral ABCD are equal.
i.e., AB = CD and AD = BD

Then, the quadrilateral can be a parallelogram, a rectangle, rhombus or a square.
In all the cases the diagonals bisects each other.

Hence, its diagonals are bisecting.

Question 13:

Fill In The Blanks

If arcs AXB and CYD of a circle are congruent, then AB : CD = ___________.

Answer 13:

Given:
Arcs AXB and CYD of a circle are congruent

We know, if any two arcs are congruent, then their corresponding chords are equal.
Thus, Chord AB = Chord CD

Hence, AB : CD = 1 : 1.

Question 14:

A, B and C are three points on a circle, then the perpendicular bisector of AB, BC and CA are ____________.

Answer 14:

Given:
A, B and C are three points on a circle

Let ABC be a triangle.
We know, a circumcentre is the point of intersection of the perpendicular bisectors of the triangle.

Thus, the perpendicular bisector of AB, BC and CA intersect at a point known as circumcentre.

Hence, the perpendicular bisector of AB, BC and CA are concurrent.

Question 15:

Fill In The Blanks

If AB and AC are equal chords of a circle, then the biesector of ∠BAC passes through the ___________.

Answer 15:

Given:
AB and AC are equal chords of a circle

Let O be the centre of the circle.

In ∆OAB and ∆OAC,
AB = AC (given)
OA = OA (common)
OB = OC (radius of the circle)

By SSS property,
∆OAB ≅ ∆OAC

Therefore, ∠OAB = ∠OAC (by C.P.C.T.)

Thus, ∠BAC = 2∠OAB.

Hence, the bisector of ∠BAC passes through the centre.

Question 16:

Fill In The Blanks

ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. If ∠CBD + ∠CDB = k ∠BAD, then k = _______.

Answer 16:

Given:
ABCD is such a quadrilateral such that A is the centre of the circle passing through B, C and D
∠CBD + ∠CDB = k∠BAD

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠CAD = 2∠CBD ...(1)
Also, ∠CAB = 2∠CDB ...(2)

Adding (1) and (2), we get
∠CAD + ∠CAB = 2∠CBD + 2∠CDB
⇒ ∠BAD = 2∠CBD + 2∠CDB
⇒ ∠BAD = 2(∠CBD + ∠CDB)
⇒ ∠CBD + ∠CDB = $\frac{1}{2}$ ∠BAD

Hence, k = $\frac{1}{2}$ .

Question 17:

Fill In The Blanks

Two chords AB and AC of a circle are on the opposite sides of the centre. If AB and AC subtend angles equal to 90^∘ and 150^∘ respectively at the centre, then ∠BAC = _________.

Answer 17:

Given:
AB and AC subtend angles equal to 90° and 150° respectively at the centre
i.e., ∠AOB = 90° and ∠AOC = 150° ...(1)

In ∆AOB,
OA = OB (radius)
∴ ∠OBA = ∠OAB (angles opposite to equal sides are equal) ...(2)

Now, ∠OBA + ∠OAB + ∠AOB = 180° (angle sum property)
⇒ 2∠OAB +  90° = 180°    (From (1) and (2))
⇒ 2∠OAB = 180° − 90°
⇒ 2∠OAB = 90°
⇒ ∠OAB = 45° ...(3)

In ∆AOC,
OA = OC (radius)
∴ ∠OCA = ∠OAC (angles opposite to equal sides are equal) ...(4)

Now, ∠OCA + ∠OAC + ∠AOC = 180° (angle sum property)
⇒ 2∠OAC +  150° = 180°    (From (1) and (4))
⇒ 2∠OAC = 180° − 150°
⇒ 2∠OAC = 30°
⇒ ∠OAC = 15° ...(5)

Thus,
∠BAC = ∠OAC + ∠OAB
⇒ ∠BAC = 15° + 45° (From (3) and (5))
⇒ ∠BAC = 60°

Hence, ∠BAC = 60°.

Question 18:

Fill In The Blanks

Two congruent circles have centres at O and O'. Arc AXb of circle centred at O, subtends an angle of 75^∘ at the centre O and arc PYQ ( or circle centred at O') subtends an angle 25^∘ at the centre O'. The ratio of the arcs AXB and PYQ is ___________.

Answer 18:

Given:
O and O' are the centres of two congruent circles
AXB of circle centred at O, subtends an angle of 75^∘ at the centre O
arc PYQ of circle centred at O' ,subtends an angle 25^∘at the centre O'

Since, the circles are congruent
Therefore, they have same radius of measure r cm. ...(1)

We know, Length of arc = $\frac{θ}{360 °} \times 2 π r$

Thus,
$Length of arc A X B = \frac{75 °}{360 °} \times 2 π r . . . (2) Length of arc P Y Q = \frac{25 °}{360 °} \times 2 π r . . . (3) \frac{Length of arc A X B}{Length of arc P Y Q} = \frac{\frac{75 °}{360 °} \times 2 π r}{\frac{25 °}{360 °} \times 2 π r} = \frac{75 °}{25 °} = \frac{3}{1}$

Hence, the ratio of the arcs AXB and PYQ is 3 : 1.

Question 19:

In the given figure, AB and CD are two equal chords of a circle with centre O. OP and OQ are perpendicular on chords AB and CD, respectively. If ∠POQ = 150^∘, then ∠APQ = __________.

Answer 19:

Given:
AB = CD
OP ⊥ AB and OQ ⊥ CD
∠POQ = 150° ...(1)

In ∆POQ,
OP = OQ (equal chords are equidistant from the centre)
∴ ∠OPQ = ∠OQP (angles opposite to equal sides are equal) ...(2)

Now, ∠OPQ + ∠OQP + ∠POQ = 180° (angle sum property)
⇒ 2∠OPQ +  150° = 180°    (From (1) and (2))
⇒ 2∠OPQ = 180° − 150°
⇒ 2∠OPQ = 30°
⇒ ∠OPQ = 15° ...(3)

Since, OP ⊥ AB
Thus, ∠OPA = 90° ....(4)

Now, ∠OPA = ∠OPQ + ∠APQ
⇒ 90° = 15° + ∠APQ    (From (3) and (4))
⇒ ∠APQ = 90° − 15°
⇒ ∠APQ = 75°

Hence, ∠APQ = 75°.

Question 20:

In the given figure, if OA = 5cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to _______.

Answer 20:

Given:
OA = 5cm ...(1)
AB = 8 cm
OD is perpendicular to AB

We know, perpendicular from the centre to the chord bisects the chord.
Therefore, AC = CB = $\frac{1}{2} A B$
⇒ AC = 4 cm ...(2)

In right angled ∆OAC,
Using pythagoras theorem
OA² = AC² + OC²
⇒ 5² = 4²+ OC² (From (1) and (2))
⇒ 25 = 16+ OC²
⇒ OC² = 25 − 16
⇒ OC² = 9
⇒ OC = 3 cm

OD = 5 cm (radius)
∴ CD = OD − OC
⇒ CD = 5 − 3
⇒ CD = 2 cm

Hence, CD is equal to 2 cm.

Page-15.112

Question 21:

In the given figure, if ∠ABC = 20∘ , then ∠AOC is equal to ____________.

Answer 21:

Given:
∠ABC = 20° ...(1)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOC = 2∠ABC
⇒ ∠AOC = 2(20°) (From (1))
⇒ ∠AOC = 40°

Hence, ∠AOC is equal to 40°.

Question 22:

In the given figure, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to ____________.

Answer 22:

Given:
AOB is a diameter of the circle
AC = BC

We know, the diameter subtends a right angle to any point on the circle.
∴ ∠ACB = 90° ...(1)

In ∆ACB,
AC = BC (given)
∴ ∠CAB = ∠CBA (angles opposite to equal sides are equal) ...(2)

Now,
∠CAB + ∠CBA + ∠ACB = 180° (angle sum property)
⇒ 2∠CAB + 90° = 180° (From (1) and (2))
⇒ 2∠CAB = 180° − 90°
⇒ 2∠CAB = 90°
⇒ ∠CAB = 45°

Hence, ∠CAB is equal to 45°.

Question 23:

In the given figure,∠AOB = 90° and ∠ABC = 30° , then ∠CAO is equal to ___________.

Answer 23:

Given:
∠AOB = 90° ...(1)
∠ABC = 30° ...(2)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOB = 2∠ACB
⇒ 90° = 2∠ACB
⇒ ∠ACB = 45° ...(3)

In ∆ACB,
∠CAB + ∠CBA + ∠ACB = 180° (angle sum property)
⇒ ∠CAB + 30° + 45° = 180° (From (2) and (3))
⇒ ∠CAB + 75°= 180°
⇒ ∠CAB = 180° −  75°
⇒ ∠CAB = 105° ...(4)

Also, in ∆OAB,
OA = OB
∴ ∠OAB = ∠OBA (angles opposite to equal sides are equal) ...(5)

Now,
∠OAB + ∠OBA + ∠AOB = 180° (angle sum property)
⇒ 2∠OAB +  90° = 180° (From (1) and (5))
⇒ 2∠OAB = 180° −  90°
⇒ 2∠OAB = 90°
⇒ ∠OAB = 45° ...(6)

∠CAO = ∠CAB − ∠OAB
= 105° − 45° (From (4) and (6))
= 60°

Hence, ∠CAO is equal to 60°.

Question 24:

In the given figure, if ∠OAB = 40∘ , then ∠ACB = ____________

Answer 24:

Given:
∠OAB = 40° ...(1)

In ∆OAB,
OA = OB
∴ ∠OAB = ∠OBA = 40° (angles opposite to equal sides are equal) ...(2)

Now,
∠OAB + ∠OBA + ∠AOB = 180° (angle sum property)
⇒ 40° + 40° + ∠AOB = 180° (From (1) and (2))
⇒ ∠AOB = 180° − 80°
⇒ ∠AOB = 100° ...(3)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOB = 2∠ACB
⇒ 100° = 2∠ACB (From (3))
⇒ ∠ACB = 50°

Hence, ∠ACB = 50°.

Question 25:

In the given figure,, if ∠DAB = 60∘ , ∠ABD = 50∘ , then ∠ACB = _________

Answer 25:

Given:
∠DAB = 60° ...(1)
∠ABD = 50° ...(2)

In ∆ADB,
∠DAB + ∠DBA + ∠ADB = 180° (angle sum property)
⇒ 60° + 50° + ∠ADB = 180° (From (1) and (2))
⇒ ∠ADB = 180° − 110°
⇒ ∠ADB = 70° ...(3)

We know, angles in the same segment of the circle are equal.
Thus, ∠ADB = ∠ACB
⇒ 70° = ∠ACB (From (3))
⇒ ∠ACB = 70°

Hence, ∠ACB = 70°.

Question 26:

In the given figure,, BC is a diameter of circle and ∠BAO = 60∘ . Then, ∠ADC = __________.

Answer 26:

Given:
BC is a diameter of circle
∠BAO = 60° ...(1)

In ∆OAB,
OA = OB
∴ ∠OAB = ∠OBA = 60° (angles opposite to equal sides are equal) ...(2)

Also,
∠AOC = ∠OAB + ∠OBA (exterior angle)
⇒ ∠AOC = 60° + 60° (From (2))
⇒ ∠AOC = 120° ...(3)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOC = 2∠ADC
⇒ 120° = 2∠ADC (From (3))
⇒ ∠ADC = 60°

Hence, ∠ADC = 60°.

Page-15.113

Question 27:

In the given figure, if AOB is a diameter and ∠ADC = 120° , then ∠CAB = ___________

Answer 27:

Given:
AOB is a diameter of circle
∠ADC = 120°

Quadrilateral ADCB is a cyclic quadrilateral.

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

Thus, ∠ADC + ∠CBA = 180°
⇒ 120° + ∠CBA = 180°
⇒ ∠CBA = 180° − 120°
⇒ ∠CBA = 60° ...(1)

We know, the diameter subtends a right angle to any point on the circle.
∴ ∠ACB = 90° ...(2)

In ∆ACB,
∠CAB + ∠CBA + ∠ACB = 180° (angle sum property)
⇒ ∠CAB + 60° + 90° = 180° (From (1) and (2))
⇒ ∠CAB = 180° − 150°
⇒ ∠CAB = 30° ...(3)

Hence, ∠CAB = 30°.

Question 28:

In the given figure, if AOC is a diameter of the circle and AXB = $\frac{1}{2}$ are BYC, then ∠BOC = __________.

Answer 28:

Given:
AOC is a diameter of circle
arc AXB = $\frac{1}{2}$ arc BYC
⇒ ∠BOA = $\frac{1}{2}$ ∠BOC ..(1)

Now, ∠BOA + ∠BOC = 180° (Angles on a straight line)
⇒ $\frac{1}{2}$ ∠BOC + ∠BOC = 180° (From (1))
⇒ $\frac{3}{2}$ ∠BOC = 180°
⇒ ∠BOC = $\frac{2}{3} \times$ 180°
⇒ ∠BOC = 120°

Hence, ∠BOC = 120°.

Question 29:

In the given figure, ∠ABC = 45^∘ , then ∠AOC = _________.

Answer 29:

Given:
∠ABC = 45^∘ ..(1)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOC = 2∠ABC
⇒ ∠AOC = 2(45°) (From (1))
⇒ ∠AOC = 90°

Hence, ∠AOC = 90°.

Question 30:

In the given figure, if ∠ADC = 130^∘ and chord BC = chord BE, then ∠CBE = _______.

Answer 30:

Given:
∠ADC = 130^∘ ...(1)
chord BC = chord BE ...(2)

Quadrilateral ADCB is a cyclic quadrilateral.

In a cyclic quadrilateral, the sum of opposite angles is 180^∘.
Thus, ∠ADC + ∠CBA = 180°
⇒ 130° + ∠CBA = 180°
⇒ ∠CBA = 180° − 130°
⇒ ∠CBA = 50° ...(3)

In ∆CBO and ∆EBO,
BC = BE (given)
OB = OB (common)
OC = OE (radius of the circle)

By SSS property,
∆OCB ≅ ∆OEB

Therefore, ∠OBC = ∠OBE = 50° (by C.P.C.T.) ...(4)

Thus, ∠CBE = ∠OBE + ∠OBC
= 50° + 50° (From (4))
= 100°

Hence, ∠CBE = 100°.

Question 31:

In the given figure, if ∠ACB = 40^∘ , then ∠AOB = ______ and ∠OAB = _____________

Answer 31:

Given:
∠ACB = 40^∘ ...(1)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOB = 2∠ACB
⇒ ∠AOB = 2(40°) (From (1))
⇒ ∠AOB = 80° ...(2)

In ∆AOB,
OA = OB (radius)
∴ ∠OAB = ∠OBA (angles opposite to equal sides are equal) ...(3)

∠OAB + ∠OBA + ∠AOB = 180° (angle sum property)
⇒ 2∠OAB + 80° = 180° (From (2) and (3))
⇒ 2∠OAB = 180° − 80°
⇒ 2∠OAB = 100°
⇒ ∠OAB = 50°

Hence, ∠AOB = 80° and ∠OAB = 50°.

Question 32:

In the given figure, AOB is a diameter of the circle and C, D , E are any three points to semi circle then ∠ACD + ∠BED = _______

Answer 32:

Given:
AOB is a diameter of the circle

We know, the diameter subtends a right angle to any point on the circle.
∴ ∠AEB = 90° ...(1)

Quadrilateral ACDE is a cyclic quadrilateral.

In a cyclic quadrilateral, the sum of opposite angles is 180^∘.
Thus, ∠ACD + ∠DEA = 180° ...(2)

Adding (1) and (2), we get
∠ACD + ∠DEA + ∠AEB = 90° + 180°
⇒ ∠ACD + ∠DEB = 270°

Hence, ∠ACD + ∠BED = 270°.

Question 33:

In the given figure, if ∠OAB = 30^∘ and ∠OCB = 57^∘ , then ∠BOC = _______ and ∠AOC = _________ .

Answer 33:

Given:
∠OAB = 30^∘ ...(1)
∠OCB = 57^∘ ...(2)

In ∆COB,
OC = OB (radius)
∴ ∠OCB = ∠OBC = 57^∘(angles opposite to equal sides are equal) ...(3)

∠OCB + ∠OBC + ∠COB = 180° (angle sum property)
⇒ 57^∘ + 57^∘ + ∠COB = 180° (From (3))
⇒ ∠COB = 180° − 114°
⇒ ∠COB = 66° ...(4)

In ∆AOB,
OA = OB (radius)
∴ ∠OAB = ∠OBA = 30^∘(angles opposite to equal sides are equal) ...(5)

∠OAB + ∠OBA + ∠AOB = 180° (angle sum property)
⇒ 30^∘ + 30^∘ + ∠AOB = 180° (From (5))
⇒ ∠AOB = 180° − 60°
⇒ ∠AOB = 120° ...(6)

Now,
∠AOB = ∠AOC + ∠COB
⇒ 120° = ∠AOC + 66° (From (4) and (6))
⇒ ∠AOC = 120° − 66°
⇒ ∠AOC = 54°

Hence, ∠BOC = 66° and ∠AOC = 54°.