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The points A , B , C ,D and E lie on a circle.
AE is paraliel to
CD.

/_EAC=70^(@);/_ABC=120^(@).
(i)
/_AEC=
qquad
(ii)
/_EDC=
qquad
(iii)
/_ECD=
qquad
(iv)
/_CED=
qquad

$6$ . $\newline$ The points A , B , C ,D and E lie on a circle. $\mathrm{AE}$ is paraliel to $\mathrm{CD}$ . $\newline$ $\angle \mathrm{EAC}=70^{\circ} ; \angle \mathrm{ABC}=120^{\circ}$ . $\newline$ (i) $\angle \mathrm{AEC}=$ $\qquad$ $\newline$ (ii) $\angle \mathrm{EDC}=$ $\qquad$ $\newline$ (iii) $\angle \mathrm{ECD}=$ $\qquad$ $\newline$ (iv) $\angle \mathrm{CED}=$ $\qquad$

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Transformations of quadratic functions

Full solution

Q.

6

.

\newline

The points A , B , C ,D and E lie on a circle.

\mathrm{AE}

is paraliel to

\mathrm{CD}

.

\newline

\angle \mathrm{EAC}=70^{\circ} ; \angle \mathrm{ABC}=120^{\circ}

.

\newline

(i)

\angle \mathrm{AEC}=

\qquad

\newline

(ii)

\angle \mathrm{EDC}=

\qquad

\newline

(iii)

\angle \mathrm{ECD}=

\qquad

\newline

(iv)

\angle \mathrm{CED}=

\qquad

Apply Alternate Interior Angles Theorem: Since $AE$ is parallel to $CD$ and angle $\angle EAC$ is given as $70$ degrees, by the Alternate Interior Angles Theorem, $\angle ECD$ is also $70$ degrees.
Calculate Angle AEC: Angle $\angle ABC$ is given as $120$ degrees. Since $AE$ is parallel to $CD$ and angle $\angle ABC$ is an exterior angle to triangle $AEC$ , $\angle AEC$ is calculated as $180 - 120 = 60$ degrees.

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