Rd Sharma 2022 for Class 9 Maths Chapter 15 - Circles

Answer:

(i) interior/exterior
(ii) concentric
(iii) the exterior
(iv) arc
(v) diameter
(vi) semi-circle
(vii) centre
(viii) three

Answer:

(i) Given that a circle is a plane figure.

As we know that a circle is a collection of those points in a plane that are at a given constant distance from a fixed point in the plane.

Thus the given statement is .

(ii) Given that line segment joining the centre to any point on the circle is a radius of the circle.

As we know that line segment joining the centre to any point on the circle is a radius of the circle.

Thus the given statement is .

(iii) Given that if a circle is divided into three equal arcs each is a major arc.

As we know that if points P, Q and R lies on the given circle C(O, r) in such a way that

Then each arc is called major arc.

Thus the given statement is .

(iv) It is given that a circle has only finite number of equal chords.

As we know that a circle having infinite number of unequal chords.

Thus the given statement is.

(v) Given that a chord of the circle, which is twice as long as its radius is diameter of the circle.

As we know that a chord of a circle which is largest to others and passing through the centre of the circle and twice as long as its radius is called diameter of the circle.

Thus the given statement is .

(vi) It is given that sector is the region between the chord and its corresponding arc.

As we know that the region between the chord and its corresponding arc is called sector.

Thus the given statement is.

(vii) Given that the degree measure of an arc is the complement of the central angle containing the arc.

As we know that the degree measure of a minor arc is the measure of the central angle containing the arc and that of a major arc is 360° minus the degree measure of the corresponding minor arc.

Let degree measure of an arc   is θ of a given circle is denoted by

Thus the given statement is.

(viii) Given that the degree measure of a semi-circle is 180°.

As we know that the diameter of a circle divides into two equal parts and each of these two arcs are known as semi-circle.

and   are semi circle

Hence,

Thus the given statement is.

Answer:

Let AB be a chord of a circle with centre O and radius 8 cm such that

AB = 12 cm

We draw and join OA.

Since, the perpendicular from the centre of a circle to a chord bisects the chord.

Now in we have

Hence the distance of chord from the centre .

Answer:

Given that OA = 10 cm and OL = 5 cm, we have to find the length of chord AB.

Let AB be a chord of a circle with centre O and radius 10 cm such that AO = 10 cm

We draw and join OA.

Since, the perpendiculars from the centre of a circle to a chord bisect the chord.

Now in we have

Hence the length of chord

Answer:

Let A, B and C are three distinct points on a circle .

Now join AB and BC and draw their perpendicular bisectors.

The point of intersection of the perpendicular bisectors is the centre of given circle.

Hence O is the centre of circle.

Answer:

Given that two different pairs of circles in the figure.

As we see that only two points A, B of first pair of circle and C, D of the second pair of circles are common points.

Thus only two points are common in each pair of circle.

Answer:

Let AB and CD be two parallel chord of the circle with centre O such that AB = 6 cm, CD = 8 cm and OP = 4 cm. let the radius of the circle be cm.

According to the question, we have to find OQ

Draw and as well as point O, Q, and P are collinear.

Let

Join OA and OC, then

OA = OC = r

Now and

So, AP = 3 cm and CQ = 4 cm

In we have

And in

Answer:

Let MN is the diameter and chord AB of circle C(O, r) then according to the question

AP = BP.

Then we have to prove that .

Join OA and OB.

       

In ΔAOP and ΔBOP

          (Radii of the same circle)

AP = BP         (P is the mid point of chord AB)

OP = OP         (Common)

Therefore,

                      (by cpct)

Hence, proved.

Answer:

Given that a line AB = 5 cm, one circle having radius of which is passing through point A and B and other circle of radius.

As we know that the largest chord of any circle is equal to the diameter of that circle.

So,

There is no possibility to draw a circle whose diameter is smaller than the length of the chord.

Answer:

Let P is the mid point of chord AB of circle C(O, r) then according to question, line OQ passes through the point P.

Then prove that OQ bisect the arc AB.

Join OA and OB.

In $△AOP \mathrm{and} △BOP$

                     (Radii of the same circle)

                     (P is the mid point of chord AB)

                     (Common)

Therefore,

                    (by cpct)

Thus

Arc AQ = arc BQ

Therefore,

Hence Proved.

Answer:

We have to prove that two different circles cannot intersect each other at more than two points.

Let the two circles intersect in three points A, B and C.

Then as we know that these three points A, B and C are non-collinear. So, a unique circle passes through these three points.

This is a contradiction to the fact that two given circles are passing through A, B, C.

Hence, two circles cannot intersect each other at more than two points.

Hence, proved.

Answer:

Let ABC be an equilateral triangle of side 9 cm and let AD be one of its medians. Let G be the centroid of. Then

We know that in an equilateral triangle centroid coincides with the circumcentre. Therefore, G is the centre of the circumcircle with circumradius GA.

As per theorem, G is the centre and . Therefore,

In we have

Therefore radius AG = $\frac{2}{3}AD=3\sqrt{3} \mathrm{cm}$

Answer:

Let AB and CD be two parallel chord of the circle with centre O such that AB = 5 cm and CD = 11 cm. let the radius of the circle be cm.

Draw and as well as point O, Q and P are collinear.

Clearly, PQ = 3 cm

Let then

In we have

…… (1)

And

…… (2)

From (1) and (2) we get

$$\begin{gathered} \Rightarrow 6x+\frac{61}{4}=\frac{121}{4} \\ \Rightarrow 6x=\frac{121-61}{4} \\ \Rightarrow 6x=\frac{60}{4} \\ \Rightarrow x=\frac{5}{2} \end{gathered}$$

Putting the value of x in (2) we get,

Answer:

Let AB and CD be two parallel chord of the circle with centre O such that AB = 5 cm, CD = 11 cm and PQ = 6 cm. Let the radius of the circle be cm.

Draw and as well as point O, Q, and P are collinear.

Clearly, PQ = 6 cm

Let OQ = x cm then

Join OA and OC, then

OA = OC = r

Nowand

So, and

In we have

…… (1)

And

…… (2)

From (1) and (2) we get

Putting the value of x in (1) we get,

Answer:

Draw perpendiculars OA and OB on RS and SM respectively.

$\mathrm{AK}=\mathrm{AL}=\frac{24}{2}=12 \mathrm{cm}$

OK = OL = OM = 20 m. (Radii of the circle)

In ΔOAK,

OA² + AK² = OK²

OA² + (12 m)² = (20 m)²

OA² = (400 − 144) m² = 256 m²

OA = 16 m

OKLM will be a kite (OK = OM and KL = LM). We know that the diagonals of a kite are perpendicular and the diagonal common to both the isosceles triangles is bisected by another diagonal.

∴∠KCL will be of 90° and KC = CM

Area of ΔORS = $\frac{1}{2}\times \mathrm{KC}\times \mathrm{OL}$
⇒ $\frac{1}{2}\times \mathrm{KC}\times \mathrm{OL}$ = $\frac{1}{2}\times 24\times 16$
⇒ $\frac{1}{2}\times \mathrm{KC}\times \mathrm{OL}$ = 192
⇒ KC × 10 = 192
⇒ KC = 19.2

Therefore, the distance between Reshma and Mandip is 38.4 m.

Answer:

It is given that AB = BC = CA

Therefore, ΔABC is an equilateral triangle.

OA (radius) = 20 m

Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC. We also know that medians intersect each other in the ratio 2 : 1. As AD is the median of the equilateral triangle ABC, we can write

$$\begin{gathered} \Rightarrow \frac{\mathrm{OA}}{\mathrm{OD}}=\frac{2}{1} \\ \Rightarrow \frac{20}{\mathrm{OD}}=\frac{2}{1} \\ \Rightarrow \mathrm{OD}=10 \mathrm{m} \end{gathered}$$

∴ AD = OA + OD = (20 + 10) m = 30 m

In ΔADC,

AC² = AD² + DC²

$$\begin{gathered} \Rightarrow {\mathrm{AC}}^2={30}^2+{\left(\frac{\mathrm{AC}}{2}\right)}^2 \\ \Rightarrow {\mathrm{AC}}^2={30}^2+\frac{1}{4}{\left(\mathrm{AC}\right)}^2 \\ \Rightarrow \frac{3}{4}{\mathrm{AC}}^2=900 \\ \Rightarrow \mathrm{AC}=20\sqrt{3} \end{gathered}$$

Therefore, the length of the string of each phone will be $20\sqrt{3}$m.

Answer:

Given that, AB and CD are two equal chords of a circle with center O. L is the mid-point of AB and M is the mid-point of CD.
(i) L is the mid-point of chord AB of the circle with center O.
$\Rightarrow \mathrm{OL}\perp \mathrm{AB}$
 $\Rightarrow \angle \mathrm{OLA}=90°$
Similarly, $\mathrm{OM}\perp \mathrm{CD}$ $\Rightarrow \angle \mathrm{OMC}=90°$
$\Rightarrow \angle \mathrm{OLA}=\angle \mathrm{OMC}$                             ....(1)

But, chord AB = chord CD
$\Rightarrow \mathrm{OL}=\mathrm{OM}$          (Equal chords are equidistant from the center)

In $∆\mathrm{OLM}$
OL = OM
$\Rightarrow \angle \mathrm{OLM}=\angle \mathrm{OML}$  (Angle opposite to equal sides are equal)      .....(2)

(ii) Subtracting (2) from (1) on both sides
$$\begin{gathered} \angle \mathrm{OLA}-\angle \mathrm{OLM}=\angle \mathrm{OMC}-\angle \mathrm{OML} \\ \Rightarrow \angle \mathrm{ALM}=\angle \mathrm{CML} \end{gathered}$$

Hence proved

 

Answer:

Given that, ∠CAB = 50°
We know that the angle subtended by the arc of a circle at the center is double the angle subtended by the arc at any point on the circumference.
$\begin{array}{rcl}\therefore \angle \mathrm{BOC}& =& 2\times \angle \mathrm{CAB}\\ & =& 2\times 50°\\ & =& 100°\end{array}$

Thus, ∠BOC = 100°

Answer:

It is given that

And (given)

We have to find

In given triangle

(Given)

 OB = OA               (Radii of the same circle)

Therefore, is an isosceles triangle.

So, $\angle OBA=\angle OAB$          ..… (1)

                       (Given)

                    [From (1)]

So

Again from figure, is given triangle and

Now in , 

                  (Radii of the same circle)

$\angle OAC=\angle OCA$

               (Given that)

Then,

Since

Hence

Answer:

We have to find in each figure.

(i) It is given that

$\angle AOC+\angle COB=180° \left[\mathrm{Linear} \mathrm{pair}\right]$

 
As we know the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Now ,$x=\frac{1}{2}\angle COB=22{\frac{1}{2}}^{°}$

Hence

(ii) As we know that = x                 [Angles in the same segment]

line is diameter passing through centre,

So,

$\angle BCA= 90° \left[\mathrm{Angle} \mathrm{inscribed} \mathrm{in} \mathrm{a} \mathrm{semicircle} \mathrm{is} \mathrm{a} \mathrm{right} \mathrm{angle} \right]$

$$\begin{gathered} \angle CAB+\angle ABC+\angle BCA=180° \left[\mathrm{Angle} \mathrm{sum} \mathrm{property}\right] \\ \Rightarrow x+40°+90°=180° \\ \Rightarrow x=50° \end{gathered}$$

(iii) It is given that

$\angle ABC=\frac{1}{2}\left(\mathrm{Reflex} \angle AOC\right)$

So

And

Then

Hence

(iv)

  (Linear pair)

And
x =
 
 
Hence,

(v) It is given that

is an isosceles triangle.
  

Therefore

And,

$$\begin{gathered} \mathrm{In} △AOB, \\ \angle AOB+\angle OBA+\angle BAO=180° \\ \Rightarrow 70°+\angle BAO=180° \\ \Rightarrow \angle BAO=110° \end{gathered}$$

$\angle AOB=2\left(\mathrm{Reflex}\angle ACB\right)$
                     

Hence,

(vi) It is given that

And

$$\begin{gathered} \angle COA+\angle AOB=180° \\ \Rightarrow \angle COA=180°-60° \\ \Rightarrow \angle COA=120° \end{gathered}$$

$∆OCA \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{triangle}.$

So

Hence,

(vii)                    (Angle in the same segment)

In we have

Hence

(viii)

As    (Radius of circle)

Therefore, is an isosceles triangle.

So          (Vertically opposite angles)

Hence,

(ix) It is given that

…… (1)          (Angle in the same segment)

$\angle ADB=\angle ACB=32°$  ......(2)           (Angle in the same segment)

Because and are on the same segment of the circle.

Now from equation (1) and (2) we have

Hence,

(x) It is given that

  
$\angle BAC=\angle BDC=35°$                (Angle in the same segment)

 Now in $∆BDC$ we have

$$\begin{gathered} \angle BDC+\angle DCB+\angle CBD=180° \\ \Rightarrow 35°+65°+\angle CBD=180° \\ \Rightarrow \angle CBD=180°-100°=80° \end{gathered}$$

Hence,

(xi)

 

                    (Angle in the same segment)

In we have

Hence

(xii)
 

          (Angle in the same segment)

is an isosceles triangle

So,                   (Radius of the same circle)

Then

Hence

Answer:

It is given that ∠ACB = 40° and ∠DPB = 120°

Construction: Join the point A and B

           (Angle in the same segment)

Now in $△BDP$ we have

$$\begin{gathered} \angle DPB+\angle PBD+\angle BDP=180° \\ \Rightarrow 120°+\angle PBD+40°=180° \\ \Rightarrow \angle PBD=20° \end{gathered}$$

Hence

Answer:

It is given that O is the centre of circle and A, B and C are points on circumference.

  (Given)

We have to find ∠ABC

The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

$$\begin{gathered} \angle ABC=\frac{1}{2}\left(\mathrm{reflex} \angle AOC\right) \\ =\frac{1}{2}\left(360°-150°\right) \\ =\frac{1}{2}\times 210° \\ =105° \end{gathered}$$

Hence,

Answer:

It is given that O is the centre and $\angle ROS=40°$

We have

In right angled triangle RQT we have

$$\begin{gathered} \angle RQT+\angle QTR+\angle TRQ=180° \\ \Rightarrow 20°+\angle QTR+90°=180° \\ \Rightarrow \angle QTR=70° \end{gathered}$$

$\mathrm{Hence}, \angle RTS=70°$

Answer:

It is given that, O is the center of circle and A, B and C are points on circumference on triangle

We have to prove that ∠x = ∠y + ∠z

∠4 and ∠3 are on same segment

So, ∠4 = ∠3           

               (Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)

…… (1)

               (Exterior angle is equal to the sum of two opposite interior angles)         …… (2)

$\angle 4=\angle z+\angle 1$               (Exterior angle is equal to the sum of two opposite interior angles)  

                 …… (3)

Adding (2) and (3)

……(4)

From equation (1) and (4) we have 

Answer:

It is given that

Two circles having center O and O' and ∠AOB = 130°

And AC is diameter of circle having center O

We have

So

Now, reflex

So

$x°=360°-230°=130°$

Hence,

Answer:

It is given that, is an equilateral triangle

We have to find

Since is an equilateral triangle.

So

And

…… (1)

So ,                                         (Sum of opposite angles of cyclic quadrilateral is

Hence

Answer:

Disclaimer: Figure given in the book was showing m∠PQR as m∠SQR. 

It is given that ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°

We have to find the m∠QSR and m∠QTR

Since ΔPQR is an isosceles triangle

So ∠PQR = ∠PRQ = 35° 

Then

Since PQTR is a cyclic quadrilateral

So

In cyclic quadrilateral QSRT we have

Hence,

and  

Answer:

It is given that O is centre of the circle and ∠BOD = 160°

We have to find the values of x and y.

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

Therefore,

Since, quadrilateral ABCD is a cyclic quadrilateral.

So,

x + y = 180°              (Sum of opposite angles of a cyclic quadrilateral is 180°.)

Hence and

Answer:

It is given that ∠BCD = 100° and ∠ABD = 70°

We have to find the ∠ADB

We have

∠A + ∠C = 180°                     (Opposite pair of angle of cyclic quadrilateral)

So,

Now in is and

Therefore,

Hence,

Answer:

It is given that, ABCD is cyclic quadrilateral in which AD || BC

We have to prove

Since, ABCD is a cyclic quadrilateral

So,

and               ..… (1)

and              (Sum of pair of consecutive interior angles is 180°) …… (2)

From equation (1) and (2) we have

…… (3)

…… (4)

Hence Proved

Answer:

It is given that, AB and CD are diameter with center O and

We have to find  

Construction: Join the point A and D to form line AD

Clearly arc AD subtends at B and at the centre.

Therefore,  $\angle AOD=2\angle ABD=100°$ …… (1)

Since CD is a straight line then
        
$\angle DOA+\angle AOC=180° \left(\mathrm{Linear} \mathrm{pair}\right)$

     

Hence

Answer:

It is given that, as diameter, is centre and

We have to find and

Since angle in a semi-circle is a right angle therefore

In we have

(Given)

    (Angle in semi-circle is right angle)

Now in we have

Hence and  

Answer:

It is given that, ABCD is a cyclic quadrilateral such that AB || CD and

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

                  ( given)

So,

Also AB || CD and BC transversal

So,

Now

Answer:

It is given that

ABCD is cyclic quadrilateral and

We have to find

Since ABCD is cyclic quadrilateral and sum of opposite pair of cyclic quadrilateral is 180°.

So

And

Therefore

Hence

Answer:

It is given that, O is the centre of the circle and $\angle DAB=50°$.

We have to find  the values of x and y.

ABCD is a cyclic quadrilateral and

So,

50° + y = 180°
y = 180° − 50°
y = 130°
 

Clearly is an isosceles triangle with OA = OB and

Then,

                        (Since)

So,

x + $\angle AOB$ = 180°     (Linear pair)

Therefore, x =  180° − 80° = 100°

Hence,

and

Answer:

It is given that, and

We have to find the

In given we have

$$\begin{gathered} \angle ABC+\angle BCA+\angle BAC=180° \left(\mathrm{Angle} \mathrm{sum} \mathrm{property}\right) \\ \Rightarrow \angle ABC=180°-\left(60°+20°\right)=100° \end{gathered}$$

In cyclic quadrilateral we have

         (Sum of pair of opposite angles of a cyclic quadilateral is 180º)

Then,

Hence

Answer:

It is given that, ABC is an equilateral triangle

We have to find and

Since is an equilateral triangle

So,

And is cyclic quadrilateral

So      (Sum of opposite pair of angles of a cyclic quadrilateral is 180°.)

Then,

Similarly BECD is also cyclic quadrilateral

So,

Hence, and  .

Answer:

It is given that, O is the centre of the circle and

We have to find the value of x, y and z.

Since, angle in the same segment are equal

So

And z = 30°               

As angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Since

Then,

y = 2z
   = 2 × 30°
   = 60°

Answer:

It is given that, and, are cyclic quadrilateral

We have to find the value of x and y.

Since , is a cyclic quadrilateral

So      (Opposite angle of a cyclic quadrilateral are supplementary)

        ()

     ..… (1)

x = 180° − 102°
   = 78°

Now in cyclic quadrilateral DCFE
x + y = 180°    (Opposite angles of a cyclic quadrilateral are supplementary)
y = 180° − 78°
   = 102°

Hence, x = 78° and y = 102°
 

Answer:

It is given that ∠A – ∠C = 60° and ABCD is a cyclic quadrilateral.

We have to prove that smaller of two is 60°

Since ABCD is a cyclic quadrilateral

So ∠A + ∠C = 180°                 (Sum of opposite pair of angles of cyclic quadrilateral is 180°)   ..… (1)

And,

∠A – ∠C = 60°               (Given)            ..… (2)

Adding equation (1) and (2) we have

So, ∠C = 60°

Hence, smaller of two is 60°.

 

Answer:

Here, ABCD is a cyclic quadrilateral, we need to find x.

In cyclic quadrilateral the sum of opposite angles is equal to 180°.

Therefore,

 $$\begin{gathered} \angle ADC+\angle ABC=180° \\ \Rightarrow 180°-80°+180°-x=180° \\ \Rightarrow x=100° \end{gathered}$$

Hence, the value of x is 100°.

Answer:

(i) It is given that , and

We have to find

In cyclic quadrilateral ABCD

      ..… (1)

      ..… (2)

Since, 

So,

Therefore in ,

So ,                   ..… (3)

Now,                   ( and is transversal)

(ii) It is given that , and

We have to find

                    _{(Angle in the same segment are equal)}

_Hence,

(iii) It is given that, ∠BCD = 100° and ∠ABD = 70°

As we know that sum of the opposite pair of angles of cyclic quadrilateral is 180°.

$$\begin{gathered} \angle DAB+\angle BCD=180° \\ \Rightarrow \angle DAB=180°-100° \\ =80° \end{gathered}$$

In ΔABD we have,

$$\begin{gathered} \angle DAB+\angle ABD+\angle BDA=180° \\ \Rightarrow \angle BDA=180°-150°=30° \end{gathered}$$

_Hence, 

Answer:

If in cyclic quadrilateral , then we have to find the other three angles.

Since, AD is parallel to BC, So,
  (Alternate interior angles)

Now, since ABCD is cyclic quadrilateral, so

And,

$\mathrm{Hence}, \angle A=110°, \angle C=70° \mathrm{and} \angle D=110°$.

Answer:

It is given that is a cyclic quadrilateral with and as its diagonals.

We have to find

Since angles in the same segment of a circle are equal

So

Since (Opposite angle of cyclic quadrilateral)

Hence

Answer:

To prove: AC = BD

Proof: We know that equal chords subtend equal at the centre of circle and the angle subtended by a chord at the centre is twice the angle subtended by it at remaining part of the circle.

$$\begin{gathered} \angle AOD=\angle BOC \left(O \mathrm{is} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{circle}\right) \\ \angle AOD=2\angle ACD \\ \mathrm{and} \angle BOC=2\angle BDC \\ \mathrm{Since}, \angle AOD=\angle BOC \\ \Rightarrow \angle ACD=\angle BDC .....\left(1\right) \\ \angle ACB=\angle ADB .....\left(2\right) \left(\mathrm{Angle} \mathrm{in} \mathrm{the} \mathrm{same} \mathrm{segment} \mathrm{are} \mathrm{equal}\right) \\ \mathrm{Adding} \left(1\right) \mathrm{and} \left(2\right) \\ \angle BCD=\angle ADC .....\left(3\right) \\ \mathrm{In} △ACD \mathrm{and} △BDC \\ CD=CD \left(\mathrm{common}\right) \\ \angle BCD=\angle ADC \left[\mathrm{Using}\left(3\right)\right] \\ AD=BC \left(\mathrm{given}\right) \\ \mathrm{Hence}, △ACD\cong BDC \left(\mathrm{SAS} \mathrm{congruency} \mathrm{criterion}\right) \\ \therefore AC=BD \left(\mathrm{cpct}\right) \\ \mathrm{Hence} \mathrm{Proved} \end{gathered}$$

Answer:

To prove: Perpendicular bisector of side AB, BC, CD and DA are concurrent i.e, passes through the same point.

Proof:

We know that the perpendicular bisector of every chord of a circle always passes through the centre.

Therefore, Perpendicular bisectors of chord AB, BC, CD and DA pass through the centre which means they all passes through the same point.

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

Answer:

Here, ABCD is a cyclic rectangle; we have to prove that the centre of the corresponding circle is the intersection of its diagonals.

Let O be the centre of the circle.

We know that the angle formed in the semicircle is 90°.

Since, ABCD is a rectangle, So

Therefore, AC and BD are diameter of the circle.

We also know that the intersection of any two diameter is the centre of the circle.

Hence, the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.

Answer:

(i) If ABCD is a cyclic quadrilateral in which AB and CD when produced meet in E such that EA =  ED, then we have to prove the following, AD || BC

(ii) EB = EC

(i) It is given that EA = ED, so

$\angle EAD=\angle EDA=x$

Since, ABCD is cyclic quadrilateral

Now,

Therefore, the adjacent angles andare supplementary

Hence, AD || BC

(ii) Since, AD and BC are parallel to each other, so,

$$\begin{gathered} \angle ECB=\angle EDA \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle EBC=\angle EAD \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \mathrm{But}, \angle EDA=\angle EAD \\ \mathrm{Therefore}, \angle ECB=\angle EBC \\ \Rightarrow EC=EB \\ \mathrm{Therefore}, △ECB \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{triangle}. \end{gathered}$$

Answer:

$$\begin{gathered} \stackrel{⏜}{QP} \mathrm{is} \mathrm{a} \mathrm{major} \mathrm{arc} \mathrm{and} \angle PSQ \mathrm{is} \mathrm{the} \mathrm{angle} \mathrm{formed} \mathrm{by} \mathrm{it} \mathrm{in} \mathrm{the} \mathrm{alternate} \mathrm{segment}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{angle} \mathrm{subtended} \mathrm{by} \mathrm{an} \mathrm{arc} \mathrm{at} \mathrm{the} \mathrm{centre} \mathrm{is} \mathrm{twice} \mathrm{the} \mathrm{angle} \mathrm{subtended} \mathrm{by} \mathrm{it} \mathrm{at} \mathrm{any} \mathrm{point} \mathrm{of} \mathrm{the} \mathrm{alternate} \mathrm{segment} \mathrm{of} \mathrm{the} \mathrm{circle}. \\ \therefore 2\angle PSQ=m\left(\stackrel{⏜}{QP}\right) \\ \Rightarrow 2\angle PSQ=360°-m\left(\stackrel{⏜}{PQ}\right) \\ \Rightarrow 2\angle PSQ=360°-\angle POQ \\ \Rightarrow 2\angle PSQ=360°-180° \left(\because \angle POQ<180°\right) \\ \Rightarrow 2\angle PSQ>180° \\ \Rightarrow \angle PSQ>90° \end{gathered}$$

Thus, the angle in a segment shorter than a semi-circle is greater than a right angle.

Answer:

$$\begin{gathered} \mathrm{To} \mathrm{prove}: \angle ABC \mathrm{is} \mathrm{an} \mathrm{acute} \mathrm{angle} \\ \mathrm{Proof}: \\ AD \mathrm{being} \mathrm{the} \mathrm{diameter} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \\ \Rightarrow \angle ACD=90° \left[\mathrm{Angle} \mathrm{in} \mathrm{a} \mathrm{semicircle} \mathrm{is} \mathrm{a} \mathrm{right} \mathrm{angle}\right] \\ \mathrm{Now}, \mathrm{in} △ACD, \angle ACD=90° \mathrm{which} \mathrm{means} \mathrm{that} \angle ADC \mathrm{is} \mathrm{an} \mathrm{acute} \mathrm{angle}. .....\left(1\right) \\ \mathrm{Again}, \angle ABC=\angle ADC \left[\mathrm{Angle} \mathrm{in} \mathrm{a} \mathrm{same} \mathrm{segment} \mathrm{are} \mathrm{always} \mathrm{equal}\right] \\ \Rightarrow \angle ABC \mathrm{is} \mathrm{also} \mathrm{an} \mathrm{acute} \mathrm{angle}. \left[\mathrm{Using}\left(1\right)\right] \\ \mathrm{Hence} \mathrm{proved} \end{gathered}$$

Answer:

We have to prove that

Let be a right angle at B and P be midpoint of AC

Draw a circle with center at P and AC diameter

Since therefore circle passing through B

So

Hence

Proved.

Answer:

Consider the smaller circle whose centre is given as ‘O’.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

Now consider the larger circle and the points ‘A’, ‘C’, ‘B’ and ‘O’ along its circumference. ‘ACBO’ form a cyclic quadrilateral.

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.

Hence ,the measure of is.

Answer:

Since both the circles are congruent, they will have equal radii. Let their radii be ‘r’.

So, from the given figure we have,

Now, since all the sides of the quadrilateral OBO’A are equal it has to be a rhombus.

One of the properties of a rhombus is that the opposite angles are equal to each other.

So, since it is given that, we can say that the angle opposite it, that is to say that should also have the same value.

Hence we get

Now, consider the first circle with the centre ‘O’ alone. ‘AB’ forms a chord and it subtends an angle of 50° with its centre, that is.

A property of a circle is that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

This means that,

Hence the measure of is

Answer:

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to .

Here we have a cyclic quadrilateral ABCD. The centre of this circle is given as ‘O’.

Since in a cyclic quadrilateral the opposite angles are supplementary, here

Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment. The angle that the chord forms with any point on the circumference of a particular segment is always the same.

Here, ‘CD’ is a chord and ‘A’ and ‘B’ are two points along the circumference on the major segment formed by the chord ‘CD’.

So,

Now,

In any triangle the sum of the interior angles need to be equal to .

Consider the triangle ΔABP,

$$\begin{gathered} \angle PAB+\angle ABP+\angle APB=180° \\ \Rightarrow \angle APB=180°-30°-58° \\ \Rightarrow \angle APB=92° \end{gathered}$$

From the figure, since ‘AC’ and ‘BD’ intersect at ‘P’ we have,

Hence the measure of is.

Answer:

Consider the given circle with the centre ‘O’. Let the radius of this circle be ‘r’. ‘AB’ forms a chord and it subtends an angle of 80° with its centre, that is.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

In any triangle the sum of the interior angles need to be equal to 180°.

Consider the triangle

Since,  , we have. So the above equation now changes to

Considering the triangle ΔABC now,

Hence, the measure of is.

Answer:

It is given that ‘ABCD’ is a parallelogram. But since ‘A’ is the centre of the circle, the lengths of ‘AB’ and ‘AD’ will both be equal to the radius of the circle.

So, we have.

Whenever a parallelogram has two adjacent sides equal then it is a rhombus.

So ‘ABCD’ is a rhombus.

Let.

We know that in a circle the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

By this property we have

In a rhombus the opposite angles are always equal to each other.

So,

Since the sum of all the internal angles in any triangle sums up to in triangle , we have

In the rhombus ‘ABCD’ since one pair of opposite angles are ‘’ the other pair of opposite angles have to be

From the figure we see that,

So now we can write the required ratio as,

Hence the ratio between the given two angles is .

Answer:

Let us first consider the triangle ΔABQ.

It is known that in a triangle the sum of all the interior angles add up to 180°.

So here in our triangle ΔABQ we have,

By a property of the circle we know that an angle formed in a semi-circle will be 90°..

In the given circle since ‘AB’ is the diameter of the circle the angle which is formed in a semi-circle will have to be 90°.

So, we have

Now considering the triangle we have,

From the given figure it can be seen that,

Now, we can also say that,

Hence the measure of the angle is 115°.

Answer:

Consider the circle with the centre ‘P’.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

Since ‘ACD’ is a straight line, we have

Now let us consider the circle with centre ‘Q’. Here let ‘E’ be any point on the circumference along the major arc ‘BD’. Now ‘CBED’ forms a cyclic quadrilateral.

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.

So here,

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, now we have

Hence, the measure of is .

Answer:

Since, O is the circumcentre of $△ABC$, So, O would be centre of the circle passing through points A, B and C.

$\angle ABC$= 90°             (Angle in the semicircle is 90°.)

$\Rightarrow \angle OAB+\angle OBC=90°$                .....(1)

As OA = OB    (Radii of the same circle)

$$\begin{gathered} \therefore \angle OAB=\angle OBA \left(\mathrm{Angle} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{sides} \mathrm{are} \mathrm{equal}\right) \\ \mathrm{or}, \angle BAC= \angle OBA \\ \mathrm{From} \left(1\right) \\ \mathit{\angle }BAC+\angle OBC=90° \end{gathered}$$

Answer:

We need to find

Answer:

We are given ABCD is a quadrilateral with center O, ∠ADE = 95° and ∠OBA = 30°

We need to find ∠OAC

We are given the following figure

Since ∠ADE = 95°

⇒ ∠ADC = 180 ° − 95° = 85°

Since squo;ABCD is cyclic quadrilateral

This means

∠ABC + ∠ADC = 180°

Since OB = OC (radius)

⇒ ∠OBC = ∠OCB = 65°

In ΔOBC

Since ÐBAC and ÐBOC are formed on the same base which is chord.

So

Consider ΔBOA which is isosceles triangle.

∠OAB = 30°

Answer:

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.

Answer:

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.

Answer:

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 = $\frac{1}{2}\left(20\right)$ =10 cm

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

Answer:

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°.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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°.

Answer:

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°.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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 = $\underline{\frac{1}{2}}$.

Answer:

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°.

Answer:

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 = $\frac{\theta }{360°}\times 2\mathrm{\pi }r$

Thus,
$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{arc} AXB=\frac{75°}{360°}\times 2\mathrm{\pi }r ...\left(2\right) \\ \mathrm{Length} \mathrm{of} \mathrm{arc} PYQ=\frac{25°}{360°}\times 2\mathrm{\pi }r ...\left(3\right) \\ \frac{\mathrm{Length} \mathrm{of} \mathrm{arc} AXB}{\mathrm{Length} \mathrm{of} \mathrm{arc} PYQ}=\frac{{\displaystyle \frac{75°}{360°}}\times 2\mathrm{\pi }r}{{\displaystyle \frac{25°}{360°}}\times 2\mathrm{\pi }r} \\ =\frac{{\displaystyle 75°}}{{\displaystyle 25°}} \\ =\frac{{\displaystyle 3}}{{\displaystyle 1}} \end{gathered}$$


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

Answer:

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°.

Answer:

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 = $\frac{1}{2}AB$
⇒ 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.

Answer:

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°.

Answer:

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°.

Answer:

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°.

Answer:

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°.

Answer:

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°.

Answer:

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°.

Answer:

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°.

Answer:

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°.

Answer:

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°.

Answer:

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°.

Answer:

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°.

Answer:

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°.

Answer:

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°.