Question

In the given figure, the straight lines AB and CD pass through the centre O of the circle. If $\angle AOD={75}^o$ and $\angle OCE={40}^o$, Then (i) $\angle CDE$ (ii) $\angle OBE$ are respectively

• A. ${75}^{o}and{55}^o.$
• B. ${50}^{o}and{25}^o.$
• C. ${40}^{o}and{55}^o.$
• D. ${40}^{o}and{75}^o.$

Answer 1

The correct option is B ${50}^{o}and{25}^o.$

$Given-ThediameterCD\&anoterextendeddiameterABintersectatthecentreO.TwolinesDB\&ABintersectatB.ThechordCEmeetsABatE.\angle AOD={75}^{o}and\angle OCE={40}^o.Tofindout-\left(i\right)\angle CDE=?\left(ii\right)\angle OBE=?Solution-DCisadiameter.So\angle DEC=Theangleinasemicircle={90}^o.\therefore \angle CDE={180}^o-(\angle OCE+\angle DEC)\left(anglesumpropertyoftriangles\right)={180}^o-({40}^o+{90}^o)={50}^o.Again\angle AOD=\angle COB={75}^{o}and\angle AOC=\angle DOB\left(bothpairsareverticallyoppositeangles\right).i.e\angle AOC+\angle DOB=2\angle DOBSo\angle AOD+\angle COB+(\angle AOC+\angle DOB)={360}^o\left(completeangle\right)⟹2\angle DOB={360}^o-(\angle AOD+\angle COB)={360}^o-({75}^o+{75}^o)={210}^o⟹\angle DOB={105}^o.Nowin\Delta ODB\angle OBE={180}^o-(\angle DOB+\angle CDE)\left(anglesumpropertyoftriangles\right)⟹\angle OBE={180}^o-({105}^o+{50}^o)={25}^o.So\angle CDEand\angle OBEarerespectively{50}^{o}and{25}^o.Ans-OptionB.$