Question Number 6681 by Tawakalitu. last updated on 11/Jul/16
Answered by Rasheed Soomro last updated on 12/Jul/16
![Join O and Y. ∵ OX=OY=OZ [Radii of same circle] ∴ △XOY and △ZOY are issoscel triangles. ∴ ∠XYO=∠OXY=30 and ∠ZYO=∠OZY=20 [Base angles of an issoscel triangle are equal] Now ∠XYZ=∠OXY+∠OZY=30+20=50 So ∠XOZ(minor)=50×2=100 [Central angle is double of inscribed angle]] Hence ∠XOZ (major (required angle) )=360−100=260](https://www.tinkutara.com/question/Q6683.png)
$${Join}\:\mathrm{O}\:{and}\:\mathrm{Y}. \\ $$$$\because\:\:\mathrm{OX}=\mathrm{OY}=\mathrm{OZ}\:\:\left[{Radii}\:{of}\:{same}\:{circle}\right] \\ $$$$\therefore\:\bigtriangleup\mathrm{XOY}\:\:{and}\:\:\bigtriangleup\mathrm{ZOY}\:{are}\:{issoscel}\:{triangles}. \\ $$$$\therefore\:\:\angle\mathrm{XYO}=\angle\mathrm{OXY}=\mathrm{30}\:\:\:{and}\:\:\:\:\angle\mathrm{ZYO}=\angle\mathrm{OZY}=\mathrm{20} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{Base}\:{angles}\:{of}\:{an}\:{issoscel}\:{triangle}\:{are}\:{equal}\right] \\ $$$${Now}\:\:\:\:\angle\mathrm{XYZ}=\angle\mathrm{OXY}+\angle\mathrm{OZY}=\mathrm{30}+\mathrm{20}=\mathrm{50} \\ $$$$\:\:\:\:\:{So}\:\:\:\:\:\angle\mathrm{XOZ}\left({minor}\right)=\mathrm{50}×\mathrm{2}=\mathrm{100} \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{Central}\:{angle}\:{is}\:{double}\:{of}\:{inscribed}\:{angle}\right]\right] \\ $$$${Hence}\:\angle\mathrm{XOZ}\:\left({major}\:\left({required}\:{angle}\right)\:\right)=\mathrm{360}−\mathrm{100}=\mathrm{260} \\ $$
Answered by Rasheed Soomro last updated on 12/Jul/16
![An other approach Join O and Y. Now in △XOY ∵ OX=OY [Radii of same circle] ∴ ∠OXY=∠OYX [Opposite angle of equal sides] But ∠OXY=30 [Given] So ∠OYX=30 [Transitive property of ′=′] ∠XOY =180−(∠OXY+ ∠OYX) =180−(30+30)=120 And in △ZOY in similar way as above ∠ZOY=140 ∠XOZ(major/required)=∠XOY +∠YOZ=120+140=260.](https://www.tinkutara.com/question/Q6684.png)
$${An}\:{other}\:{approach} \\ $$$${Join}\:{O}\:{and}\:{Y}. \\ $$$${Now}\:{in}\:\bigtriangleup{XOY} \\ $$$$\:\:\:\:\:\:\:\:\because\:\:\:\:{OX}={OY}\:\:\:\:\:\:\left[{Radii}\:{of}\:{same}\:{circle}\right] \\ $$$$\:\:\:\:\:\:\:\therefore\:\:\:\:\:\angle{OXY}=\angle{OYX}\:\:\:\left[{Opposite}\:{angle}\:{of}\:{equal}\:{sides}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{But}\:\:\angle{OXY}=\mathrm{30}\:\:\:\:\:\left[{Given}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{So}\:\:\angle{OYX}=\mathrm{30}\:\:\left[{Transitive}\:{property}\:{of}\:'='\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\angle{XOY}\:=\mathrm{180}−\left(\angle{OXY}+\:\angle{OYX}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{180}−\left(\mathrm{30}+\mathrm{30}\right)=\mathrm{120} \\ $$$${And}\:{in}\:\bigtriangleup{ZOY}\:{in}\:{similar}\:{way}\:{as}\:{above} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\angle{ZOY}=\mathrm{140} \\ $$$$ \\ $$$$\angle{XOZ}\left({major}/{required}\right)=\angle{XOY}\:+\angle{YOZ}=\mathrm{120}+\mathrm{140}=\mathrm{260}. \\ $$
Commented by Tawakalitu. last updated on 12/Jul/16
$${Thanks}\:{so}\:{much}. \\ $$