AB is a chord of a circle with centre P. Poin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $AB$ is a chord of a circle with centre $P$. Point $C$ is a point on the major arc $AB$.According to the circle theorem,the angle subte

Vedclass · Accepted Answer

$45^{\circ}$ and $90^{\circ}$

Vedclass · Answer

(A) Given that $AB$ is a chord of a circle with centre $P$. Point $C$ is a point on the major arc $AB$.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,$\angle APB = 2 \angle ACB$.Given that $\angle ACB + \angle APB = 135^{\circ}$.Substituting $\angle APB = 2 \angle ACB$ in the given equation:$\angle ACB + 2 \angle ACB = 135^{\circ}$$3 \angle ACB = 135^{\circ}$$\angle ACB = \frac{135^{\circ}}{3} = 45^{\circ}$.Now,calculating $\angle APB$:$\angle APB = 2 	imes 45^{\circ} = 90^{\circ}$.Thus,$\angle ACB = 45^{\circ}$ and $\angle APB = 90^{\circ}$.

$AB$ is a chord of a circle with centre $P$. Point $C$ is a point other than $A$ and $B$ on the major arc $AB$. If $\angle ACB + \angle APB = 135^{\circ}$,then find $\angle ACB$ and $\angle APB$.

Similar Questions

$O$ is the circumcentre of the triangle $ABC$ and $D$ is the mid-point of the base $BC$. Prove that $\angle BOD = \angle A$.

$P$ is the centre of the circle of radius $20\, cm$. $AB$ is a chord of the circle. If $AB = 32\, cm$,then find the distance of the chord $AB$ from the centre $P$.

In a circle with centre $O$,$PQ$ and $XY$ are chords. If $\angle POQ = 120^{\circ}$,$\angle OXY = 30^{\circ}$ and $XY = 8 \, cm$,then $PQ$ is ........... $cm$.

In a circle with centre $P$,the length of a radius is $\sqrt{2} \ cm$. It is divided into two segments by chord $AB$ of length $2 \ cm$. If $C$ is a point on the major arc $AB$,then find the value of $\angle ACB$. (in $^{\circ}$)

$ABCD$ is a quadrilateral such that $A$ is the centre of the circle passing through $B, C$ and $D$. Prove that $\angle CBD + \angle CDB = \frac{1}{2} \angle BAD$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module