In a circle with centre P,AB is a chord and p | vedclass

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(D) Let the angle subtended by the chord $AB$ at the centre be $\angle APB = 	heta$. According to the theorem,the angle subtended by an arc at the ce

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(D) Let the angle subtended by the chord $AB$ at the centre be $\angle APB = 	heta$. 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,$\angle APB = 2 	imes \angle ACB$,which implies $\angle ACB = \frac{	heta}{2}$. Given that $\angle ACB + \angle APB = 150^{\circ}$. Substituting the values,we get $\frac{	heta}{2} + 	heta = 150^{\circ}$. $\frac{3	heta}{2} = 150^{\circ}$. $	heta = 150^{\circ} 	imes \frac{2}{3} = 100^{\circ}$. Thus,$\angle APB = 100^{\circ}$.

In a circle with centre $P$,$AB$ is a chord and point $C$ is a point other than $A$ and $B$ on the major arc $AB$. If $\angle ACB + \angle APB = 150^{\circ}$,then find $\angle APB$. (in $^{\circ}$)

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Write True or False and justify your answer.
The angles subtended by a chord at any two points of a circle are equal.

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