$ \mathrm{ABCD} $ is a cyclic quadrilateral whose diagonals intersect at a point $ \mathrm{E} $. If $ \angle \mathrm{DBC}=70^{\circ} $, $ \angle \mathrm{BAC} $ is $ 30^{\circ} $, find $ \angle \mathrm{BCD} $. Further, if $ \mathrm{AB}=\mathrm{BC} $, find $ \angle \mathrm{ECD} $.

AcademicMathematicsNCERTClass 9

Given:

\( \mathrm{ABCD} \) is a cyclic quadrilateral whose diagonals intersect at a point \( \mathrm{E} \). 

\( \angle \mathrm{DBC}=70^{\circ} \), \( \angle \mathrm{BAC} \) is \( 30^{\circ} \)

\( \mathrm{AB}=\mathrm{BC} \)
To do:

We have to find \( \angle \mathrm{BCD} \) and \( \angle \mathrm{ECD} \).

Solution:


We know that,

The angles in the same segment of a circle are equal.

This implies,

$\angle BDC = \angle BAC=30^o$

In $\triangle BCD$,

$\angle BDC + \angle DBC + \angle BCD = 180^o$

$30^o+70^o+\angle BCD=180^o$

$100^o+\angle BCD=180^o$

$\angle BCD=180^o-100^o$

$\angle BCD=80^o$

$AB = BC$

This implies,

$\angle BCA = \angle BAC= 80^o$      (Angles opposite to equal sides in a triangle are equal)

$\angle ECD = \angle BCD - \angle BCA$

$= 80^o - 30^o$

$= 50^o$

Hence, $\angle BCD = 80^o$ and $\angle ECD = 50^o$.

raja
Updated on 10-Oct-2022 13:46:42

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