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. \[\Rightarrow \] \[\angle X+\angle Y+\angle Z+\angle O={{360}^{o}}\] \[\Rightarrow \] \[\theta +\theta +\alpha +\alpha +\angle O={{360}^{o}}\] \[\Rightarrow \] \[\angle O={{360}^{o}}-2(\alpha +\theta )\] \[\Rightarrow \] \[\angle O={{360}^{o}}-2(\angle OZX+\angle XZY+\angle OXZ+\angle ZXY)\] \[\Rightarrow \]\[\angle O={{360}^{o}}-2\,(\angle OZX+\angle OXZ)\] \[-2(\angle XZY+\angle ZXY)\] \[\Rightarrow \] \[\angle O={{360}^{o}}-2(\angle OZX+\angle XZY+\angle OXZ+\angle ZXY)\]\[\Rightarrow \]\[\angle O={{360}^{o}}-{{360}^{o}}+2\angle O-2(\angle XZY+\angle ZXY)\] \[\Rightarrow \]\[\angle O=2\,(\angle XYZ+\angle ZXY)\]You need to login to perform this action.
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