Concise Mathematics Class 10 ICSE Solutions for Chapter 17 - Circle

Exercise-17A

Long Answer Type

1. In the given figure O is the centre of the circle. ∠AOB and ∠OCB are ${30}^{\circ }$ and ${40}^{\circ }$ respectively. Find AOC. Show your steps of working.

Ans: On joining the points A and C we get the triangles $\mathrm{\Delta }AOC$ and $\mathrm{\Delta }ABC.$

In $\mathrm{\Delta }AOC$,

$\mathrm{\angle }OAC=x\text{and}\mathrm{\angle }OCA=x$ (∵ OA=OC (radius))

$\mathrm{\angle }AOC={180}^{\circ }-(\mathrm{\angle }OAC+\mathrm{\angle }OCA)$

$\mathrm{\angle }AOC={180}^{\circ }-2x$------(1)

 In $\mathrm{\Delta }ABC$

∠BAC=${30}^{\circ }+x$

$\mathrm{\angle }BCA={40}^{\circ }+x$

We know that in a triangle sum of angles is ${180}^{\circ }$

$\mathrm{\angle }ABC={180}^{\circ }-(\mathrm{\angle }BAC+\mathrm{\angle }BCA)$

$={180}^{\circ }-({30}^{\circ }+x+{40}^{\circ }+x)$

$={110}^{\circ }-2x$

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

$\mathrm{\angle }AOC=2\mathrm{\angle }ABC$

$\Rightarrow {180}^{\circ }-2x=2({110}^{\circ }-2x)$

$\Rightarrow 4x-2x={220}^{\circ }-180$

$\Rightarrow 2x={40}^{\circ }$

$\Rightarrow x={20}^{\circ }$

So from eq.1 the value of $\mathrm{\angle }AOC$ is:

$\mathrm{\angle }AOC={180}^{\circ }-2x={180}^{\circ }-{40}^{\circ }={140}^{\circ }$

2. In the given figure, $\mathrm{\angle }BAD={65}^{\circ }$, $\mathrm{\angle }ABD={70}^{\circ },\mathrm{\angle }BDC={45}^{\circ }$

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

Ans:  In $\mathrm{\Delta }ADB$

$\mathrm{\angle }ADB+\mathrm{\angle }DAB+\mathrm{\angle }ABD={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ADB+{65}^{\circ }+{70}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ADB={180}^{\circ }-{135}^{\circ }={45}^{\circ }$

Since, $\mathrm{\angle }ADC=\mathrm{\angle }ADB+\mathrm{\angle }BDC$

$\mathrm{\angle }ADC={45}^{\circ }+{45}^{\circ }={90}^{\circ }$

$\mathrm{\angle }ADC={90}^{\circ }$

Hence the triangle $\mathrm{\Delta }ADC$is a right angled triangle and we know that when a triangle is inserted in a circle in such a way that the triangle is a right triangle then one of the sides of the triangle is the diameter of the circle, Hence AC is the diameter of the triangle.

(ii). Find $\mathrm{\angle }ACB$ 

Ans: We know that angles in the same segments of circle are equal. 

$\mathrm{\angle }ACB=\mathrm{\angle }ADB$ (∵$\mathrm{\angle }ADB={45}^{\circ }$)

Hence, $\mathrm{\angle }ACB={45}^{\circ }$

3. Given O is the centre of the circle and $AOB={70}^{\circ }$, Calculate the value of :

(i).$\mathrm{\angle }OCA$

Ans: We know that the angle at the centre is double 

the angle at the circumference subtended by the same chord.

$\mathrm{\angle }AOB=2\mathrm{\angle }ACB$

So the value of $\mathrm{\angle }ACB={\displaystyle \frac{{70}^{\circ }}{2}}={35}^{\circ }$

ACB and OAC are b/w the same line segments.

Hence $\mathrm{\angle }ACB=OCA={35}^{\circ }$------(1)

(ii). $\mathrm{\angle }OAC$

Ans: ∵ OA=OC (radii of circle)

Thus the values

 $\mathrm{\angle }OCA=\mathrm{\angle }OAC={35}^{\circ }$[from eq.1]

4. In each of the following figures, O is the centre of the circle. Find the values of a,b and c.

(i).

Ans: The angle at the centre is double the angle at the circumference subtended by the same chord.

$2b={130}^{\circ }$

$\Rightarrow b={\displaystyle \frac{1}{2}}\times {130}^{\circ }={65}^{\circ }$

By the theorem, the opposite angles of a cyclic quadrilateral are supplementary.

Hence $a+b={180}^{\circ }$

$\Rightarrow a+{65}^{\circ }={180}^{\circ }$

$\Rightarrow a={180}^{\circ }-{65}^{\circ }={115}^{\circ }$

Thus the values of a and b are respectively ${115}^{\circ }\mathrm{\&}{65}^{\circ }$.

(ii)

Ans: Given, central angle =${112}^{\circ }$

Reflex angle of  ${112}^{\circ }$= ${360}^{\circ }-{112}^{\circ }={248}^{\circ }$

We know that the angle at the centre is double the angle at the circumference by the same arc.

Hence, $2c=$ Reflex angle of ${112}^{\circ }$

$2c={248}^{\circ }$

$\Rightarrow c={\displaystyle \frac{{248}^{\circ }}{2}}={124}^{\circ }$

Thus the value of angle c is ${124}^{\circ }$.

5. In each of the following figures, O is the centre of the circle. Find the values of a,b,c and d.

(i).

Ans: Here O is the centre of the circle and BD is the Diameter of the given circle and side of the triangle ABD as well.

We know the angle at the semicircle is a right angle triangle. Hence $\mathrm{\angle }DAB={90}^{\circ }$

The sum of angles in a triangle is ${180}^{\circ }$

$\mathrm{\angle }DAB+\mathrm{\angle }ADB+\mathrm{\angle }DBA={180}^{\circ }$

$\Rightarrow {90}^{\circ }+\mathrm{\angle }ADB+{35}^{\circ }={180}^{\circ }$  (Given  $\mathrm{\angle }DBA={35}^{\circ }$)

$\Rightarrow \mathrm{\angle }ADB={180}^{\circ }-{125}^{\circ }={55}^{\circ }$

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

$\mathrm{\angle }ADB=\mathrm{\angle }ACB$

$a={55}^{\circ }$ ($\mathrm{\angle }ADB={55}^{\circ }$ and $\mathrm{\angle }ACB=a$)

Thus the value of a is ${55}^{\circ }$.

(ii).

Ans: In a circle angles at the circumference subtended by the same chord are always equal. Angles $\mathrm{\angle }DAC$ and $\mathrm{\angle }DBC$ are subtended by the chord DC.

$\mathrm{\angle }DAC=\mathrm{\angle }DBC={25}^{\circ }$

By the exterior angle property the exterior angle in a triangle is equal to the sum of opposite interior angles.

$b+{25}^{\circ }={120}^{\circ }$

$\Rightarrow b={95}^{\circ }$

Thus the value of $b$ is ${95}^{\circ }$

(iii).

Ans: We know that the angle at the centre is double the angle at the circumference subtended by the same chord.

$\mathrm{\angle }AOB=2.\mathrm{\angle }ACB$

$\Rightarrow \mathrm{\angle }AOB=2\times {50}^{\circ }={100}^{\circ }$

Also, OA=OB (Radii of the circle)

Hence the $\mathrm{\Delta }AOB$ is an isosceles triangle.

So, $\mathrm{\angle }OAB=\mathrm{\angle }OBA=c$

The sum of angles in a triangle is always ${180}^{\circ }$

$\mathrm{\angle }OAB+\mathrm{\angle }AOB+\mathrm{\angle }OBA={180}^{\circ }$

$\Rightarrow c+\mathrm{\angle }AOB+c={180}^{\circ }$

$\Rightarrow 2c={180}^{\circ }-{{100}^{\circ }}^{}$

$\Rightarrow c={\displaystyle \frac{{80}^{\circ }}{2}}={40}^{\circ }$

Thus the value of c is ${40}^{\circ }$.

(IV).

Ans: Given, angle $\mathrm{\angle }PBA={45}^{\circ }$

AB is the diameter of the circle hence $\mathrm{\Delta }APB$ is a right angle triangle and we know that angle in the semicircle is a right triangle. So ($\mathrm{\angle }APB={90}^{\circ }$).

The sum of angles in a triangle is always ${180}^{\circ }$

$\mathrm{\angle }PAB+\mathrm{\angle }PBA+\mathrm{\angle }APB={180}^{\circ }$

$\Rightarrow \mathrm{\angle }PAB+{45}^{\circ }+{90}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }PAB={180}^{\circ }-{135}^{\circ }$

$\Rightarrow \mathrm{\angle }PAB={45}^{\circ }$---------(1)

Now, the angles subtended by the same chord are equal. Here the angels $\mathrm{\angle }PAB$ and $\mathrm{\angle }PCB$ subtended by the chord PB. 

Hence $\mathrm{\angle }PAB=\mathrm{\angle }PCB$

From eq. 1

$\mathrm{\angle }PCB=\mathrm{\angle }PAB={45}^{\circ }$

$d={45}^{\circ }$   ($\because \mathrm{\angle }PCB$=d)

Thus the value of angle d is ${45}^{\circ }$.

6. In the figure, AB is the common chord of the two circles. If AC and AD are diameters; prove that D,B and C are in a straight line. $O_1$and $O_2$ are the centres of two circles.

Ans:

Given, $O_1$and $O_2$ are the centres of the circles.

Here AD and AC are the diameter of the circles. So the angles $\mathrm{\angle }DBA$ and $\mathrm{\angle }ABC$are in the semicircles and we know that the angle in the semicircle is a right angle.

Hence, $\mathrm{\angle }DBA=\mathrm{\angle }CBA={90}^{\circ }$

On Adding both the angles 

$\mathrm{\angle }DBA+\mathrm{\angle }CBA={90}^{\circ }+{90}^{\circ }={180}^{\circ }$

Thus D,B and C form a straight line.

7. In the figure, given below, find:

$\left(i\right)\mathrm{\angle }BCD,$

(ii). $\mathrm{\angle }ADC,$

(iii). $\mathrm{\angle }ABC$.

Show steps of your working. 

(i). $\mathrm{\angle }BCD$

Ans: In a cyclic quadrilateral the sum of 

opposite angles are always ${180}^{\circ }$.

$\mathrm{\angle }DAB+\mathrm{\angle }BCD={180}^{\circ }$

Given, $\mathrm{\angle }DAB={105}^{\circ }$ 

Now, ${105}^{\circ }+\mathrm{\angle }BCD={180}^{\circ }$

$\Rightarrow \mathrm{\angle }BCD={180}^{\circ }-{105}^{\circ }$

$\Rightarrow \mathrm{\angle }BCD={75}^{\circ }$

Thus the value of $\mathrm{\angle }BCD$ is ${75}^{\circ }$.

(ii). $\mathrm{\angle }ADC$

Ans: From the figure, $AB||CD$, and we know that the sum of interior angles on the same side of parallel lines is ${180}^{\circ }$.

$\mathrm{\angle }BAD+\mathrm{\angle }ADC={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ADC={180}^{\circ }-{105}^{\circ }={75}^{\circ }$

Thus the value of $\mathrm{\angle }ABC$is ${75}^{\circ }$.

(iii). $\mathrm{\angle }ABC$

Ans: In a cyclic quadrilateral the sum of opposite angles is always ${180}^{\circ }$.

$\mathrm{\angle }ADC+\mathrm{\angle }ABC={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ABC={180}^{\circ }-{75}^{\circ }={105}^{\circ }$

Thus the value of $\mathrm{\angle }ABC$ is ${105}^{\circ }$.

8. In the given figure, O is the centre of the circle. If $\mathrm{\angle }AOB={140}^{\circ }$ and $\mathrm{\angle }OAC={50}^{\circ }$, find:

(i). $\mathrm{\angle }ACB$

Ans: Given, $\mathrm{\angle }AOB={140}^{\circ }$and $\mathrm{\angle }OAC={50}^{\circ }$

Reflex angle of $\mathrm{\angle }AOB$ is ${360}^{\circ }-{140}^{\circ }={220}^{\circ }$

We know that the angle at the centre is double the angle at the circumference.

Reflex angle of $\mathrm{\angle }AOB=2\mathrm{\angle }ACB$

$\Rightarrow \mathrm{\angle }ACB={\displaystyle \frac{1}{2}}\times \text{Reflex angle of}\mathrm{\angle }AOB$

$\Rightarrow \mathrm{\angle }ACB={\displaystyle \frac{1}{2}}\times {220}^{\circ }={{110}^{\circ }}^{}$

Thus the angle $\mathrm{\angle }ACB$ is ${110}^{\circ }$.

(ii).$\mathrm{\angle }OAB,$

Ans: Since OA=OB (Radii of circle)

Hence $\mathrm{\angle }OAB=\mathrm{\angle }OBA$---------(1)

And we know that the sum of angles of a triangle 

is always ${180}^{\circ }.$

So, $\mathrm{\angle }OAB+\mathrm{\angle }AOB+\mathrm{\angle }OBA={180}^{\circ }$

From eq. 1

$2\mathrm{\angle }OBA={180}^{\circ }-{140}^{\circ }={40}^{\circ }$

$\Rightarrow \mathrm{\angle }OBA={20}^{\circ }$

∵$\mathrm{\angle }OAB=\mathrm{\angle }OBA$

Só, $\mathrm{\angle }OAB=\mathrm{\angle }OBA={20}^{\circ }$

Thus the value of $\mathrm{\angle }OAB$ is ${20}^{\circ }$.

(iii).$\mathrm{\angle }OBC$

Ans: In $\mathrm{\Delta }ACB$

$\mathrm{\angle }ACB+\mathrm{\angle }CBA+\mathrm{\angle }CAB={180}^{\circ }$

$\because \mathrm{\angle }CAB={50}^{\circ }-\mathrm{\angle }OAB={50}^{\circ }-{20}^{\circ }={30}^{\circ }$

$\Rightarrow {110}^{\circ }+\mathrm{\angle }CBA+{30}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }CBA={180}^{\circ }-{140}^{\circ }={40}^{\circ }$------(2)

Since, $\mathrm{\angle }OBC=\mathrm{\angle }OBA+\mathrm{\angle }CBA={40}^{\circ }+{20}^{\circ }={60}^{\circ }$

(iv).$\mathrm{\angle }CBA,$

Ans: In $\mathrm{\Delta }ABC,$

From eq. 2 the value of $\mathrm{\angle }CBA$ is ${40}^{\circ }$.

9. Calculate: 

(i).$\mathrm{\angle }CDB,$

Ans: We know that the angle subtended by the same chord on the circle is equal.

Here in the given circle $\mathrm{\angle }CDB$ and $\mathrm{\angle }BAC$ are subtended by the chord BC.

Hence $\mathrm{\angle }CDB=\mathrm{\angle }BAC$

$\because \mathrm{\angle }BAC={49}^{\circ }$

So, $\mathrm{\angle }CDB={49}^{\circ }$

Thus the value of $\mathrm{\angle }CDB$ is ${49}^{\circ }$.

(ii).$\mathrm{\angle }ABC,$

Ans: The angle subtended by the same chord on the circle is equal.

Here in the given circle $\mathrm{\angle }ADC$ and $\mathrm{\angle }ABC$ are subtended by the chord AC.

Hence $\mathrm{\angle }ADC=\mathrm{\angle }ABC$

$\because \mathrm{\angle }ADC={43}^{\circ }$

So, $\mathrm{\angle }ABC={43}^{\circ }$

Thus the value of $\mathrm{\angle }ABC$ is ${43}^{\circ }$.

(iii).$\mathrm{\angle }ACB$

Ans: By the angle sum property the sum of angles in a triangle is ${180}^{\circ }$.

In $\mathrm{\Delta }ACB$,

$\mathrm{\angle }ACB+\mathrm{\angle }BAC+\mathrm{\angle }ABC={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ACB+{49}^{\circ }+{43}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ACB={180}^{\circ }-{92}^{\circ }={88}^{\circ }$

Thus the value of $\mathrm{\angle }ACB$ is ${88}^{\circ }$.

10. In the figure, given below, ABCD is a cyclic quadrilateral in which $\mathrm{\angle }BAD={75}^{\circ }$; $\mathrm{\angle }ABD={58}^{\circ }$ and $\mathrm{\angle }ADC={77}^{\circ }$. Find:

(i). $\mathrm{\angle }BDC,$

Ans: By the angle sum property we know that the sum 

of all interior angles in a triangle is ${180}^{\circ }$.

In $\mathrm{\Delta }ADB$

$\mathrm{\angle }ADB+\mathrm{\angle }ABD+\mathrm{\angle }BAD={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ADB+{58}^{\circ }+{75}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ADB={180}^{\circ }-{133}^{\circ }={47}^{\circ }$

$\because \mathrm{\angle }ADC=\mathrm{\angle }ADB+\mathrm{\angle }BDC$

$\Rightarrow {77}^{\circ }={47}^{\circ }+\mathrm{\angle }BDC$  (Given $\mathrm{\angle }ADC={{77}^{\circ }}^{}$)

$\mathrm{\angle }BDC={77}^{\circ }-{47}^{\circ }={30}^{\circ }$

Thus the value of $\mathrm{\angle }BDC$ is ${30}^{\circ }$.

(ii). $\mathrm{\angle }BCD,$

Ans: By the property of cyclic quadrilaterals, the opposite angles of a cyclic quadrilateral are supplementary. Here $\mathrm{\angle }BAD$ and $\mathrm{\angle }BCD$ are opposite angles. 

So, $\mathrm{\angle }BAD+\mathrm{\angle }BCD={180}^{\circ }$

$\Rightarrow {75}^{\circ }+\mathrm{\angle }BCD={180}^{\circ }$ (From the figure, $\mathrm{\angle }BAD={75}^{\circ }$)

$\Rightarrow \mathrm{\angle }BCD={180}^{\circ }-{75}^{\circ }={105}^{\circ }$

Thus the value of $\mathrm{\angle }BCD$ is ${105}^{\circ }$.

(iii). $\mathrm{\angle }BCA$

Ans: Angles subtended by the same chord in the same segment of a circle are always equal. Here the angels  $\mathrm{\angle }BCA$ and $\mathrm{\angle }ADB$ are subtended by the chord AB.

Hence, $\mathrm{\angle }BCA=\mathrm{\angle }ADB$ 

$\because \mathrm{\angle }ADB={47}^{\circ }$

Then, $\mathrm{\angle }BCA={47}^{\circ }$

Thus the value of $\mathrm{\angle }BCA$ is ${47}^{\circ }$.

11. In the following figure, O is the centre of the circle and $\mathrm{\Delta }ABC$ is equilateral.

Find: 

(i). $\mathrm{\angle }ADB$

Ans: Given that $\mathrm{\Delta }ABC$ is an equilateral triangle.

So all the angles will be equal.

So, In$\mathrm{\Delta }ABC$

$\mathrm{\angle }ACB=\mathrm{\angle }ABC=\mathrm{\angle }CAB$

$\Rightarrow \mathrm{\angle }ACB+\mathrm{\angle }ABC+\mathrm{\angle }CAB={180}^{\circ }$

$\Rightarrow 3\mathrm{\angle }ACB={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ACB={60}^{\circ }$

We know that angles subtended by the same chord are always equal. Here $\mathrm{\angle }ACB$ and $\mathrm{\angle }ADB$ are subtended by the chord AB.

Hence, $\mathrm{\angle }ADB=\mathrm{\angle }ACB={60}^{\circ }$

Thus the value of $\mathrm{\angle }ADB$ is ${60}^{\circ }$.

(ii).$\mathrm{\angle }AEB$

Ans: Join OA and OB.

The angle at the centre is double the angle at the circumference of the circle by the same chord.

$\mathrm{\angle }AOB=2\mathrm{\angle }ACB$

$\because \mathrm{\angle }ACB={60}^{\circ }$

∴$\mathrm{\angle }AOB=2\times {60}^{\circ }={120}^{\circ }$

Here the reflex angle of $\mathrm{\angle }AOB$ will be double the angle $\mathrm{\angle }AEB$ as these are subtended by the same chord AB.

Hence, Reflex angle of $\mathrm{\angle }AOB=2\mathrm{\angle }AEB$

$\Rightarrow \mathrm{\angle }AEB={\displaystyle \frac{1}{2}}\times Reflexangleof\mathrm{\angle }AOB$

$\Rightarrow \mathrm{\angle }AEB={\displaystyle \frac{1}{2}}\times ({360}^{\circ }-{120}^{\circ })={\displaystyle \frac{1}{2}}\times {240}^{\circ }={120}^{\circ }$

Thus the value of $\mathrm{\angle }AEB$ is ${120}^{\circ }$.

12. Given: $\mathrm{\angle }CAB={75}^{\circ }$ and $\mathrm{\angle }CBA={50}^{\circ }.$ Find the value of $\mathrm{\angle }DAB+\mathrm{\angle }ABD.$

Ans: By the angle sum property the sum of angles in a triangle is always ${180}^{\circ }.$

In $\mathrm{\Delta }ABC$,

$\mathrm{\angle }CAB+\mathrm{\angle }ABC+\mathrm{\angle }ACB={180}^{\circ }$

$\Rightarrow {75}^{\circ }+{50}^{\circ }+\mathrm{\angle }ACB={180}^{\circ }$[Given $\mathrm{\angle }CAB={75}^{\circ }\mathrm{\&}\mathrm{\angle }CBA={50}^{\circ }]$

$\Rightarrow \mathrm{\angle }ACB={180}^{\circ }-{125}^{\circ }={55}^{\circ }$

The angle subtended by the same chord is always equal.

Here $\mathrm{\angle }ACB$and $\mathrm{\angle }ADB$ are subtended by the chord AB.

Hence, $\mathrm{\angle }ADB=\mathrm{\angle }ACB={55}^{\circ }$

In $\mathrm{\Delta }ADB$,

$\mathrm{\angle }DAB+\mathrm{\angle }ABD+\mathrm{\angle }ADB={180}^{\circ }$

$\Rightarrow \mathrm{\angle }DAB+\mathrm{\angle }ABD={180}^{\circ }-{55}^{\circ }={125}^{\circ }$

Therefore the value of $\mathrm{\angle }DAB+\mathrm{\angle }ABD$ is ${125}^{\circ }$.

13. ABCD is a cyclic quadrilateral in a circle with centre O. If $\mathrm{\angle }ADC={130}^{\circ },$ find $\mathrm{\angle }BAC.$

Ans: We know that the angle made in a semicircle is the right angle.

Here AB is a diameter and $\mathrm{\angle }ACB$ will be the right angle.

So,  $\mathrm{\angle }ACB={90}^{\circ }$

In a cyclic quadrilateral, opposite angles are supplementary.

So, $\mathrm{\angle }ADC+\mathrm{\angle }ABC={180}^{\circ }$

$\Rightarrow \mathrm{\angle }ABC={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Now, By using angle sum property in $\mathrm{\Delta }ABC$

$\mathrm{\angle }ABC+\mathrm{\angle }BAC+\mathrm{\angle }ACB={180}^{\circ }$

$\Rightarrow {50}^{\circ }+\mathrm{\angle }BAC+{90}^{\circ }={180}^{\circ }$ ($\mathrm{\angle }ABC={50}^{\circ },\mathrm{\angle }ABC={90}^{\circ }$)

$\Rightarrow \mathrm{\angle }BAC={180}^{\circ }-{140}^{\circ }={40}^{\circ }$

Thus the value of angle $\mathrm{\angle }BAC$is ${40}^{\circ }$.

14. In the figure, given alongside, AOB is a diameter of the circle and $\mathrm{\angle }AOC={110}^{\circ },$ find $\mathrm{\angle }BDC$.

Ans: First we join points A and D.

In a circle the angle at the centre is double the angle at the circumference subtended by the same chord.

$\mathrm{\angle }AOC=2\mathrm{\angle }ADC$

$\Rightarrow \mathrm{\angle }ADC={\displaystyle \frac{1}{2}}\times {110}^{\circ }={55}^{\circ }$

Now, Angle made in the semicircle is a right angle. Here angle $\mathrm{\angle }ADB$ is the angle subtended by the diameter AB so this is the angle made in semicircle. Hence $\mathrm{\angle }ADB$ is the right angle.

Therefore, $\mathrm{\angle }ADB={90}^{\circ }$

Since, $\mathrm{\angle }ADB=\mathrm{\angle }ADC+\mathrm{\angle }BDC$

$\Rightarrow {90}^{\circ }={55}^{\circ }+\mathrm{\angle }BDC$

$\Rightarrow \mathrm{\angle }BDC={35}^{\circ }$

Thus the value of angle $\mathrm{\angle }BDC$ is ${35}^{\circ }$.

15. In the following figure, O is the centre of the circle, $\mathrm{\angle }AOB={60}^{\circ }$ and $\mathrm{\angle }BDC={100}^{\circ }$. Find $\mathrm{\angle }OBC$.

Ans: In a circle the angle at the centre is double the angle at the circumference by the same chord. Here angle $\mathrm{\angle }AOB$ is the angle at centre and $\mathrm{\angle }ACB$ at circumference subtended by the same arc AB. 

Hence, $\mathrm{\angle }AOB=2\mathrm{\angle }ACB$

$\Rightarrow \mathrm{\angle }ACB={\displaystyle \frac{1}{2}}\times \mathrm{\angle }AOB={\displaystyle \frac{1}{2}}\times {60}^{\circ }={30}^{\circ }$

In $\mathrm{\Delta }BOC$ , By using the angle sum property

$\Rightarrow \mathrm{\angle }BOC+\mathrm{\angle }OBC+\mathrm{\angle }OCB={180}^{\circ }$

$\Rightarrow {100}^{\circ }+\mathrm{\angle }OBC+{30}^{\circ }={180}^{\circ }$  (∵ $\mathrm{\angle }OCB=\mathrm{\angle }ACB$)

$\Rightarrow \mathrm{\angle }OBC={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Thus the value of $\mathrm{\angle }OBC$ is ${50}^{\circ }$.

16. In cyclic quadrilateral ABCD, $\mathrm{\angle }DAC={27}^{\circ };\mathrm{\angle }DBA={50}^{\circ }$ and $\mathrm{\angle }ADB={33}^{\circ }.$ Calculate:

(i). $\mathrm{\angle }DBC,$

Ans:  In a circle, Angles at the circumference subtended by the same chord are equal. Here the $\mathrm{\angle }DAC$ and $\mathrm{\angle }DBC$ are subtended by the chord DC.

Hence $\mathrm{\angle }DAC=\mathrm{\angle }DBC$

$\Rightarrow \mathrm{\angle }DBC={27}^{\circ }$  (∵$\mathrm{\angle }DAC={27}^{\circ }$)

So the value of $\mathrm{\angle }DBC$ is ${27}^{\circ }$

(ii). $\mathrm{\angle }DCB,$

Ans: In a circle, Angles at the circumference subtended by the same chord are equal.

$\mathrm{\angle }DCA=\mathrm{\angle }DBA={50}^{\circ }-----\left(1\right)$(Subtended by the same chord AD)

$\mathrm{\angle }ACB=\mathrm{\angle }ADB={33}^{\circ }------\left(2\right)$  (Subtended by the same chord AB)

From the figure, 

$\mathrm{\angle }DCB=\mathrm{\angle }DCA+\mathrm{\angle }ACB$

From eq. 1 and eq.2 $\mathrm{\angle }DCA={50}^{\circ }\mathrm{\&}\mathrm{\angle }ACB={33}^{\circ }$

$\mathrm{\angle }DCB={50}^{\circ }+{33}^{\circ }={{83}^{\circ }}^{}$

So the value of $\mathrm{\angle }DCB$is ${{83}^{\circ }}^{}$.

(iii). $\mathrm{\angle }CAB.$

Ans: In a cyclic quadrilateral the sum of opposite angles is ${180}^{\circ }$.

$\mathrm{\angle }DCB+\mathrm{\angle }DAB={180}^{\circ }$

$\Rightarrow \mathrm{\angle }DAB={180}^{\circ }-83={97}^{\circ }$

$\Rightarrow \mathrm{\angle }DAC+\mathrm{\angle }CAB={97}^{\circ }$  (∵$\mathrm{\angle }DAB=\mathrm{\angle }DAC+\mathrm{\angle }CAB$)

$\Rightarrow \mathrm{\angle }CAB={97}^{\circ }-{27}^{\circ }={70}^{\circ }$

Thus the value of $\mathrm{\angle }CAB$ is ${70}^{\circ }$.

17. In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If $\mathrm{\angle }AOC={80}^{\circ }$ and $\mathrm{\angle }CDE={40}^{\circ },$ find the number of degrees in:

(i). $\mathrm{\angle }DCE$

Ans: The angle made in a semicircle is the right angle. Here CD is the diameter so the angle $\mathrm{\angle }CED$ will be at the right angle. Hence $\mathrm{\angle }CED={90}^{\circ }$.

Now, By using the angle sum property in $\mathrm{\Delta }CDE$.

$\mathrm{\angle }CED+\mathrm{\angle }CDE+\mathrm{\angle }DCE={180}^{\circ }$

$\Rightarrow {90}^{\circ }+{40}^{\circ }+\mathrm{\angle }DCE={180}^{\circ }$

$\Rightarrow \mathrm{\angle }DCE={180}^{\circ }-{130}^{\circ }={50}^{\circ }$

Thus the value of $\mathrm{\angle }DCE$ is ${50}^{\circ }$.

(ii). $\mathrm{\angle }ABC$

Ans: We know that In a triangle the exterior angle is equal to the sum of a pair of opposite angles.

In $\mathrm{\Delta }BOC$,

$\mathrm{\angle }AOC$ is the exterior angle.

$\mathrm{\angle }AOC=\mathrm{\angle }OCB+\mathrm{\angle }OBC$

$\Rightarrow {80}^{\circ }={50}^{\circ }+\mathrm{\angle }OBC$

$\Rightarrow \mathrm{\angle }OBC={30}^{\circ }$

Since, $\mathrm{\angle }ABC=\mathrm{\angle }OBC={30}^{\circ }$

Thus the value of $\mathrm{\angle }ABC$ is ${30}^{\circ }$.

18. In the given figure, AC is the diameter of a circle,whose centre is O. A circle is described on AO as diameter. AE, is a chord of the larger circle, intersects the smaller circle at B. Prove that AB=BE.

Ans: First we join points B and O.

Now In triangle $\mathrm{\Delta }ABO$ The side OA is also the diameter of a smaller circle. Hence the $\mathrm{\Delta }ABO$ is a right angle triangle. And the angle $\mathrm{\angle }ABO={90}^{\circ }$.

So, $AB\mathrm{\perp }OB\Rightarrow AE\mathrm{\perp }OB$

And we know that if we draw the perpendicular on the chord from the centre then the perpendicular bisects the chord.

Hence, AB=BE.

19. (a). In the following figure,

(i). If $\mathrm{\angle }BAD={96}^{\circ }$, find$\mathrm{\angle }BCD$and $\mathrm{\angle }BFE.$

Ans: ABCD is a cyclic quadrilateral and we know that the sum of opposite angles in a cyclic quadrilateral is ${180}^{\circ }$.

So, $\mathrm{\angle }BAD+\mathrm{\angle }BCD={180}^{\circ }$

$\Rightarrow {96}^{\circ }+\mathrm{\angle }BCD={180}^{\circ }$

$\Rightarrow \mathrm{\angle }BCD={180}^{\circ }-{96}^{\circ }={84}^{\circ }$

Now,

$\mathrm{\angle }BCD+\mathrm{\angle }BCE={180}^{\circ }$

$\Rightarrow \mathrm{\angle }BCE={180}^{\circ }-{84}^{\circ }={96}^{\circ }$

Here BCEF is also a cyclic quadrilateral.

So, $\mathrm{\angle }BCE+\mathrm{\angle }BFE={180}^{\circ }$

$\Rightarrow \mathrm{\angle }BFE={180}^{\circ }-{96}^{\circ }={84}^{\circ }$

Thus the value of $\mathrm{\angle }BCD$ and $\mathrm{\angle }BFE$ are respectively ${96}^{\circ }$ and ${84}^{\circ }$.

(ii). Prove that AD is parallel to FE.

Ans: ADEF is a cyclic quadrilateral 

Here $\mathrm{\angle }BAD+\mathrm{\angle }BFE={96}^{\circ }+{84}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }BAD+\mathrm{\angle }BFE={180}^{\circ }$

Here $\mathrm{\angle }BAD$ and $\mathrm{\angle }BFE$ are co-interior angles on the same side of a pair of lines AD and EF. 

By converse of the co-interior angle theorem, if a transversal line intersects two lines in such a way that the sum of a pair of co-interior angles is ${180}^{\circ }$then the lines are parallel lines.

Hence, AD||EF.

(b). ABCD is a parallelogram. A circle through vertices A and B meets side BC at point P and side AD at point Q. Show that quadrilateral PCDQ is cyclic.

Ans: ABPQ is cyclic quadrilateral.

$\mathrm{\angle }PCD=\mathrm{\angle }BAQ$-----(1) (opposite angles of parallelogram)

$\mathrm{\angle }BAQ=\mathrm{\angle }QPC$-------(2) (exterior angle of cyclic Quadrilateral)

From eq.(1)&(2)

$\mathrm{\angle }PCD=\mathrm{\angle }QPC$------(3)

Here ABCD is a parallelogram, So the sum of the co-interior angles on the same side is ${180}^{\circ }$.

Hence, $\mathrm{\angle }QDC+\mathrm{\angle }PCD={180}^{\circ }$

From eq.(3) $\mathrm{\angle }PCD=\mathrm{\angle }QPC$

$\Rightarrow \mathrm{\angle }QDC+\mathrm{\angle }QPC={180}^{\circ }$

We know that if the sum of opposite angles of a quadrilateral is ${180}^{\circ }$ then the quadrilateral is a cyclic quadrilateral. Here $\mathrm{\angle }QDC\mathrm{\&}\mathrm{\angle }QPC$ are the opposite angles of quadrilateral PCDQ and we got the sum as ${180}^{\circ }$. 

Therefore PCDQ is a cyclic quadrilateral.

20. Prove that:

(i)The parallelogram inside the circle is a rectangle.

Ans: 

Let ABCD is a parallelogram inside the circle.

In parallelogram, Opposite sides and opposite angles are equal.

Therefore, AB=CD and $\mathrm{\angle }BAD=\mathrm{\angle }BCD$

Since, ABCD is also a cyclic quadrilateral.

So, $\mathrm{\angle }BAD+\mathrm{\angle }BCD={180}^{\circ }$

$\Rightarrow 2\mathrm{\angle }BAD={180}^{\circ }$

$\Rightarrow \mathrm{\angle }BAD={90}^{\circ }$

Therefore, $\mathrm{\angle }BAD=\mathrm{\angle }BCD={90}^{\circ }$

Similarly, The other two angles will be right angles and the other pair of opposite sides are equal. Hence all the respective sides are equal. 

$\mathrm{\angle }BAD=\mathrm{\angle }BCD=\mathrm{\angle }ABC=\mathrm{\angle }ADC={90}^{\circ }$

AB=CD and  BC=AD

So the parallelogram ABCD is a rectangle.

(ii). The rhombus, inscribed in a circle, is a square.

Ans: In a rhombus opposite angles are always equal.

Hence, $\mathrm{\angle }BAD=\mathrm{\angle }BCD$

Here ABCD is also a cyclic quadrilateral.

So, $\mathrm{\angle }BAD+\mathrm{\angle }BCD={180}^{\circ }$

$\Rightarrow 2\mathrm{\angle }BAD={180}^{\circ }$ ($\because \mathrm{\angle }BAD=\mathrm{\angle }BCD$)

$\Rightarrow \mathrm{\angle }BAD={90}^{\circ }$

∴$\mathrm{\angle }BAD=\mathrm{\angle }BCD={90}^{\circ }$

Similarly the other two angles are right angle and equal.

So, $\mathrm{\angle }BAD=\mathrm{\angle }BCD=\mathrm{\angle }ADC=\mathrm{\angle }ABC={90}^{\circ }$ and AB=BC=CD=AD.

Therefore the quadrilateral ABCD is a square.

21. In the given figure, AB=AC. Prove that DECB is an isosceles trapezium.

Ans: Given AB=AC------(1)

At first we join the points D and E.

∵ AB=AC

Therefore $\mathrm{\angle }ABC=\mathrm{\angle }ACB\Rightarrow \mathrm{\angle }DBC=\mathrm{\angle }ECB$

Here DBEC is a cyclic quadrilateral.

So, $\mathrm{\angle }DEC+\mathrm{\angle }DBC={180}^{\circ }$

$\Rightarrow \mathrm{\angle }DEC+\mathrm{\angle }ECB={180}^{\circ }$

So, DE||BC and given AB=AC

∴ $\mathrm{\angle }AED=\mathrm{\angle }ACB$ and $\mathrm{\angle }ADE=\mathrm{\angle }ABC$   (Corresponding angles)

Hence, $\mathrm{\angle }AED=\mathrm{\angle }ADE$

Therefore the sides AD=AE-----(2)

From eq.(1)-eq.(2)

$AB-AD=AC-AE$

$\Rightarrow BD=CE$

$\because DE||BC$ and $BD=CE$

Therefore DECB is an isosceles trapezium.

22. Two circles intersect at P and Q. Through P diameter PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.

Ans: 

Let O and O’are the centres of the two intersecting circles and PA and PB are the diameters. Here Angles $\mathrm{\angle }PQA$ and $\mathrm{\angle }PQB$ are subtended by the diameters PA and PB respectively, which means these angles are made in semicircle and we know that if in a circle an angle is made in semicircle then that angle is right angle.

So, $PQA$and$PQB$are right angles.

$\mathrm{\angle }PQA={90}^{\circ }$and $\mathrm{\angle }PQB={90}^{\circ }$

Now on adding the above angles,

$\mathrm{\angle }PQA+\mathrm{\angle }PQB={90}^{\circ }+{90}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }PQA+\mathrm{\angle }PQB={180}^{\circ }$

Thus AQB is a straight line. Hence point A,Q and b are collinear points.

23. The figure given below, shows a circle with centre O.

Given: $\mathrm{\angle }AOC=a$ and $\mathrm{\angle }ABC=b.$

(i). Find the relationship between a and b.

Ans: We know that in a circle the angle at the centre is double the angle at any point on circumference subtended by the same chord.

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