Question

In the given figure, $AB$ is a diameter of a circle with centre $O$. If $ADE$ and $CBE$ are straight lines, meeting a $E$ such that $\angle BAD={35}^o$ and $\angle BED={25}^o$, Then : (i) $\angle DCB$ (ii) $\angle DBC$ (iii) $\angle BDC$, are respectively.

• A. ${55}^o,{100}^o\&{35}^o$
• B. ${25}^o,{120}^o\&{35}^o$
• C. ${35}^o,{115}^o\&{30}^o$
• D. ${65}^o,{135}^o\&{255}^o$

Answer 1

The correct option is C ${35}^o,{115}^o\&{30}^o$

$Given-OisthecentreofacirclewithdiameterAB.AD\&BCarechordswhichhavebeenextendedtointersectatE.AB\&CDarejoined.\angle BAD={35}^o\&\angle BED={25}^o.Tofindout-\angle DCB,\angle DBC\&\angle BDCSolution-\angle BAD=\angle DCB={35}^o......\left(i\right)\left[sincebothofthemhavebeensubtendedbythechordBDtothecircumferenceatA\&C\right].Now\angle ACB={90}^{o}since\angle ACBisangleinasemicircle,ABbeingthediameter.\therefore \angle ACD=\angle ACB-\angle DCB={90}^o-{35}^o={55}^o.Again\angle ABD=\angle ACD={55}^o\left[sincebothofthemhavebeensubtendedbythechordADtothecircumferenceatB\&C\right].Nowin\Delta ACE,wehave\angle CAD={180}^o-(\angle ACB+\angle BEA)={180}^o-({90}^o+{25}^o)={65}^o.\therefore \angle BAC=\angle CAD-\angle BAD={65}^o-{35}^o={30}^o.\therefore \angle BDC=\angle BAC={30}^o.......\left(ii\right)\left[sincebothofthemhavebeensubtendedbythechordBCtothecircumferenceatA\&D\right].AgainACBDisacyclicquadrilateral.\therefore \angle CAD+\angle DBC={180}^o\left(Thesumoftheoppositeanglesofacyclicquadrilateralis{180}^o\right).So\angle DBC={180}^o-\angle CAD={180}^o-{65}^o={115}^o.......\left(iii\right).\therefore \angle DCB,\angle DBC\&\angle BDCare{35}^o,{115}^o\&{30}^{o}respectively.Hence,optionCiscorrect.$