Question

In the figure, $O$ is the centre of the circle. If $\angle BCO={30}^o,$ find $x$ and $y$.

• A. $x={60}^o$ & $y={45}^o$
• B. $x={30}^o$ & $y={15}^o.$
• C. $x={15}^o$ & $y={45}^o$
• D. $x={20}^o$ & $y={15}^o$

Answer 1

The correct option is B $x={30}^o$ & $y={15}^o.$

$Given-BA,BD\&BCarethechordsofthecirclewithcentreO.AE\perp BC\&OD\perp AE.\angle OCE={30}^o.Tofindout-\angle BAE=x=?And\angle DBC=y=?Solution-WejoinAD\&DC.\angle ABD=\frac{1}{2}\angle AOD=\frac{1}{2}\times {90}^o={45}^o.(\because Theangleatthecentreofacirclesubtendedbyachordisdoubleofthatatthecircumference.).\therefore In\Delta OECwehave\angle COE={180}^o-(\angle OEC+\angle OCE)={180}^o-({90}^o+{30}^o)={60}^o.\left(anglesumpropertyoftriangles\right)\therefore \angle DOC=\angle DOE-\angle COE={90}^o-{60}^o={30}^o.Now\angle DBC=y=\frac{1}{2}\angle DOC=\frac{1}{2}\times {30}^o={15}^o.(\because Theangleatthecentreofacirclesubtendedbyachordisdoubleofthatatthecircumference.).Again\angle ABE=y+{45}^o={45}^o+{15}^o={60}^o.\therefore In\Delta ABEwehave\angle BAE=x={180}^o-(\angle ABE+\angle AEB)={180}^o-({60}^o+{90}^o)={30}^o.\left(anglesumpropertyoftriangles\right)Sox={30}^o\&y={15}^o.Ans-OptionB.$