question_answer

In the given figure, O is the centre of the circle, then \[\angle XOZ\] is _____.

A) \[2\angle XZY\]            

B)        \[2\angle Y\]             

C)        \[2\angle Z\]

D)        \[2(\angle XZY+\angle YXZ)\]

Correct Answer: D

Solution :

\[OX\text{ }=OY=OZ=r\]           (radius of circle) Taking \[\angle X=\theta \]  and \[\angle Z=\alpha \] Then in \[\Delta YXO,\text{ }OX=OY\Rightarrow \angle X=\angle OYX=\theta \] Now, in \[\Delta OZY,\,OZ=OY\Rightarrow \angle Z=\angle OYZ=\alpha \] Now, XOZY is a quadrilateral.                \[\Rightarrow \]  \[\angle X+\angle Y+\angle Z+\angle O={{360}^{o}}\] \[\Rightarrow \]  \[\theta +\theta +\alpha +\alpha +\angle O={{360}^{o}}\] \[\Rightarrow \] \[\angle O={{360}^{o}}-2(\alpha +\theta )\] \[\Rightarrow \] \[\angle O={{360}^{o}}-2(\angle OZX+\angle XZY+\angle OXZ+\angle ZXY)\] \[\Rightarrow \]\[\angle O={{360}^{o}}-2\,(\angle OZX+\angle OXZ)\]                         \[-2(\angle XZY+\angle ZXY)\] \[\Rightarrow \] \[\angle O={{360}^{o}}-2(\angle OZX+\angle XZY+\angle OXZ+\angle ZXY)\]\[\Rightarrow \]\[\angle O={{360}^{o}}-{{360}^{o}}+2\angle O-2(\angle XZY+\angle ZXY)\] \[\Rightarrow \]\[\angle O=2\,(\angle XYZ+\angle ZXY)\]