Question

If the figure, AB is parallel to DC, $\angle BCE={80}^o$ and $\angle BAC={25}^o$. Then (i) $\angle CAD$ (ii) $\angle CBD$ (iii) $\angle ADC$
are respectively

• A. (i) $\angle CAD={100}^o$
  (ii) $\angle CBD={45}^o$
  (iii) $\angle ADC={65}^o$
• B. (i) $\angle CAD={65}^o$
  (ii) $\angle CBD={45}^o$
  (iii) $\angle ADC={55}^o$
• C. (i) $\angle CAD={45}^o$
  (ii) $\angle CBD={65}^o$
  (iii) $\angle ADC={90}^o$
• D. (i) $\angle CAD={55}^o$
  (ii) $\angle CBD={55}^o$
  (iii) $\angle ADC={100}^o$

Answer 1

The correct option is D (i) $\angle CAD={55}^o$

(ii) $\angle CBD={55}^o$
(iii) $\angle ADC={100}^o$
$Given-ABCDisacyclicquadrilateral.AC\&BDarediagonals.\angle BAC={25}^{o}and\angle BCE={80}^o.Tofindout-\left(i\right)\angle CAD=?\left(iI\right)\angle CBD=?\left(i\right)\angle ADC=?Solution-\angle BCE+\angle BCD=1{80}^o\left(linearpair\right)⟹\angle BCD=1{80}^o-\angle BCE=1{80}^o-{80}^o=1{00}^o\left(i\right)AlsoAB\parallel CDi.e\angle CAB=\angle ACD={25}^o.\left(alternativeangles\right).........\left(ii\right)\therefore \angle ACB=\angle BCD-\angle ACD=1{00}^o-{25}^o={75}^o.SincethechordABsubtends\angle ACB\&\angle ADBtothecircumferenceofthegivencircleatC\&Drespectively,weget\angle ACB=\angle ADB={75}^o(\because Theangles,subtendedbyachordofacircletodifferentpointsofthecircumfereceofthesamecircle,areequal).AgainBCsubtends\angle CAB\&\angle CDBtothecircumferenceofthegivencircleatA\&Drespectively,weget\angle CAB=\angle CDB={25}^o(\because theangles,subtendedbyachordofacircletodifferentpointsofthecircumfereceofthesamecircle,areequal).\therefore \angle ADC=\angle ADB+\angle CDB={75}^o+{25}^o={100}^o.....\left(ansiii\right).NowABCDisacyclicquadrilateral.\therefore \angle BCD+\angle BAD={180}^o⟹\angle BAD={180}^o-\angle BCD={180}^o-{100}^o={80}^o(fromeqn.i)⟹\angle BAC+\angle CAD={80}^o⟹\angle CAD={80}^o-\angle BAC={80}^o-{25}^o={55}^o\left(ansi\right).AgainDCsubtends\angle CAD\&\angle CBDtothecircumferenceofthegivencircleatA\&Brespectively,weget\angle CAD=\angle CBD={55}^o(\because theangles,subtendedbyachordofacircletodifferentpointsofthecircumfereceofthesamecircle,areequal).\therefore \angle ADC=\angle ADB+\angle CDB={75}^o+{25}^o={100}^o.....\left(ansii\right).So\left(i\right)\angle CAD={55}^o,\left(ii\right)\angle CBD={55}^o,\left(iii\right)\angle ADC={100}^o.AnsOptionD.$