Angle Subtended by an Arc of a Circle Questions in English

(A) Let the chord be $AB$ and the center of the circle be $O$. Since the chord $AB$ is equal to the radius of the circle,we have $OA = OB = AB$.
Therefore,$\Delta AOB$ is an equilateral triangle.
Since each angle of an equilateral triangle is $60^{\circ}$,we have $\angle AOB = 60^{\circ}$.
The reflex angle subtended by the chord at the center is $\text{reflex } \angle AOB = 360^{\circ} - 60^{\circ} = 300^{\circ}$.
The angle subtended by the chord at a point $C$ on the minor arc is half of the reflex angle subtended at the center: $\angle ACB = \frac{1}{2} \times 300^{\circ} = 150^{\circ}$.
The angle subtended by the chord at a point $D$ on the major arc is half of the angle subtended at the center: $\angle ADB = \frac{1}{2} \times 60^{\circ} = 30^{\circ}$.
Thus,the angle subtended by the chord on the minor arc is $150^{\circ}$ and on the major arc is $30^{\circ}$.

(N/A) Given a triangle $ABC$ inscribed in a circle,where the angle bisectors of $\angle A, \angle B$,and $\angle C$ intersect the circumcircle at $D, E$,and $F$ respectively.
Join $DE, EF$,and $FD$.
Since angles subtended by the same arc in the same segment are equal:
$\angle FDA = \angle FCA = \frac{1}{2} \angle C$ (as $CF$ is the bisector of $\angle C$)
$\angle EDA = \angle EBA = \frac{1}{2} \angle B$ (as $BE$ is the bisector of $\angle B$)
Adding these two equations:
$\angle FDE = \angle FDA + \angle EDA = \frac{1}{2} \angle C + \frac{1}{2} \angle B = \frac{1}{2}(\angle B + \angle C)$
Since $\angle A + \angle B + \angle C = 180^{\circ}$,we have $\angle B + \angle C = 180^{\circ} - \angle A$.
Therefore,$\angle FDE = \frac{1}{2}(180^{\circ} - \angle A) = 90^{\circ} - \frac{1}{2} \angle A$.
Similarly,$\angle FED = 90^{\circ} - \frac{1}{2} \angle B$ and $\angle EFD = 90^{\circ} - \frac{1}{2} \angle C$.
Thus,the angles of $\Delta DEF$ are $90^{\circ} - \frac{A}{2}, 90^{\circ} - \frac{B}{2}$,and $90^{\circ} - \frac{C}{2}$.

(D) In $\Delta OAB$,
$OA = OB$ (Radii of the same circle).
Therefore,$\angle OAB = \angle OBA = 40^{\circ}$ (Angles opposite to equal sides are equal).
In $\Delta OAB$,the sum of angles is $180^{\circ}$.
So,$\angle AOB = 180^{\circ} - (40^{\circ} + 40^{\circ}) = 180^{\circ} - 80^{\circ} = 100^{\circ}$.
According to the theorem,the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 100^{\circ} = 50^{\circ}$.

(C) In $\Delta OAB$,we have:
$OA = OB$ [Radii of the same circle]
Therefore,$\angle ABO = \angle BAO$ [Angles opposite to equal sides are equal].
Given $\angle BAO = 60^{\circ}$,so $\angle ABO = 60^{\circ}$.
In $\Delta OAB$,the sum of angles is $180^{\circ}$,so $\angle AOB = 180^{\circ} - (60^{\circ} + 60^{\circ}) = 60^{\circ}$.
Since $BC$ is a diameter,$\angle AOC = 180^{\circ} - \angle AOB = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Thus,$\angle ADC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 120^{\circ} = 60^{\circ}$.

(B) The statement is $False$.
Justification: According to the theorem of circles,angles subtended by a chord at any two points in the same segment of a circle are equal.
However,if the two points lie in different segments (one in the major segment and one in the minor segment),the angles subtended by the chord at these points are supplementary,not equal. Therefore,the statement is incorrect as it does not specify that the points must lie in the same segment.

(N/A) We know that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle AOC = 2 \times \angle ABC$.
Given that $\angle ABC = 45^{\circ}$.
Substituting the value,we get $\angle AOC = 2 \times 45^{\circ} = 90^{\circ}$.
Since $\angle AOC = 90^{\circ}$,the lines $OA$ and $OC$ are perpendicular to each other.
Hence,$OA \perp OC$ is proved.

$(50^{\circ})$ Since the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle,we have:
$\angle AOB = 2 \angle ACB = 2 \times 40^{\circ} = 80^{\circ}$
In $\Delta OAB$,since $OA = OB$ (radii of the same circle),the angles opposite to equal sides are equal. Therefore,$\angle OAB = \angle OBA = p^{\circ}$.
Using the angle sum property in $\Delta OAB$:
$p^{\circ} + p^{\circ} + \angle AOB = 180^{\circ}$
$2p^{\circ} + 80^{\circ} = 180^{\circ}$
$2p^{\circ} = 180^{\circ} - 80^{\circ}$
$2p^{\circ} = 100^{\circ}$
$p^{\circ} = 50^{\circ}$
Hence,$\angle OAB = 50^{\circ}$.

(N/A) Let the circle have center $O$ and radius $r = \sqrt{2} \text{ cm}$. Let $BC$ be the chord of length $2 \text{ cm}$.
Join $OB$ and $OC$. In $\triangle OBC$,we have $OB = OC = \sqrt{2} \text{ cm}$ and $BC = 2 \text{ cm}$.
Check if $\triangle OBC$ is a right-angled triangle:
$OB^2 + OC^2 = (\sqrt{2})^2 + (\sqrt{2})^2 = 2 + 2 = 4$.
$BC^2 = (2)^2 = 4$.
Since $OB^2 + OC^2 = BC^2$,by the converse of the Pythagoras theorem,$\angle BOC = 90^{\circ}$.
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle BAC = \frac{1}{2} \angle BOC = \frac{1}{2} \times 90^{\circ} = 45^{\circ}$.
Hence,it is proved that the angle subtended by the chord in the major segment is $45^{\circ}$.

(C) According to the circle theorem,the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Here,the arc $AB$ subtends $\angle APB$ at the centre $P$ and $\angle ACB$ at the remaining part of the circle.
Therefore,$\angle APB = 2 \angle ACB$.
Given that $\angle ACB = 50^{\circ}$.
Substituting the value,we get $\angle APB = 2 \times 50^{\circ} = 100^{\circ}$.
Thus,the value of $\angle APB$ is $100^{\circ}$.

(D) Given that $AB$ is a chord of a circle with centre $P$. Point $C$ is a point on the major arc $AB$.
We are given that $\angle APB = 130^{\circ}$.
According to the circle theorem,the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle APB = 2 \angle ACB$.
Substituting the given value:
$130^{\circ} = 2 \angle ACB$
$\angle ACB = \frac{130^{\circ}}{2} = 65^{\circ}$.
Thus,$\angle ACB = 65^{\circ}$.

(B) Given: Radius $r = \sqrt{2} \ cm$ and chord length $AB = 2 \ cm$.
In $\triangle APB$,$AP = BP = r = \sqrt{2} \ cm$ and $AB = 2 \ cm$.
Check if $\triangle APB$ is a right-angled triangle: $AP^2 + BP^2 = (\sqrt{2})^2 + (\sqrt{2})^2 = 2 + 2 = 4$.
Since $AB^2 = 2^2 = 4$,we have $AP^2 + BP^2 = AB^2$.
By the converse of the Pythagoras theorem,$\triangle APB$ is a right-angled triangle with $\angle APB = 90^{\circ}$.
The angle subtended by the chord $AB$ at the centre is $\angle APB = 90^{\circ}$.
The angle subtended by the same chord at any point $C$ on the major arc is half the angle subtended at the centre.
Therefore,$\angle ACB = \frac{1}{2} \angle APB = \frac{1}{2} \times 90^{\circ} = 45^{\circ}$.

(D) According to the circle theorem,the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Here,the arc $AB$ subtends $\angle APB$ at the centre $P$ and $\angle ACB$ at the point $C$ on the major arc.
Therefore,$\angle APB = 2 \times \angle ACB$.
Given $\angle ACB = 47^{\circ}$.
So,$\angle APB = 2 \times 47^{\circ} = 94^{\circ}$.

(A) piece of a circle between two points on a circle is known as an $\operatorname{arc}$.
An $\operatorname{arc}$ is a portion of the circumference of a circle.

(D) Let the angle subtended by the chord $AB$ at the centre be $\angle APB = \theta$.
According to the theorem,the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle APB = 2 \times \angle ACB$,which implies $\angle ACB = \frac{\theta}{2}$.
Given that $\angle ACB + \angle APB = 150^{\circ}$.
Substituting the values,we get $\frac{\theta}{2} + \theta = 150^{\circ}$.
$\frac{3\theta}{2} = 150^{\circ}$.
$\theta = 150^{\circ} \times \frac{2}{3} = 100^{\circ}$.
Thus,$\angle APB = 100^{\circ}$.