$4$ . In the diagram, $\mathrm{O}$ is the centre of the circle, $\angle \mathrm{OCD}=40^{\circ}$ and $\angle \mathrm{ADC}=29^{\circ}$ . CE intersects $D A$ and $O A$ at $Q$ and $R$ respectively and $O C$ intersects DA at $P$ . Find $\newline$ a. $\angle \mathrm{AEC} \quad[2]$ $\newline$ b. $\angle \mathrm{OCD}=40^{\circ}$ $0$ $\newline$ c. $\angle \mathrm{OCD}=40^{\circ}$ $1$ $\newline$ d. $\angle \mathrm{OCD}=40^{\circ}$ $2$

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Algebra 2

Transformations of quadratic functions

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4

. In the diagram,

\mathrm{O}

is the centre of the circle,

\angle \mathrm{OCD}=40^{\circ}

and

\angle \mathrm{ADC}=29^{\circ}

. CE intersects

D A

and

O A

Q

and

R

respectively and

O C

intersects DA at

P

. Find

\newline

\angle \mathrm{AEC} \quad[2]

\newline

\angle \mathrm{OCD}=40^{\circ}

0

\newline

\angle \mathrm{OCD}=40^{\circ}

1

\newline

\angle \mathrm{OCD}=40^{\circ}

2

Find Angle AEC: a. To find $\angle AEC$ , we know that $\angle OCD$ is $40$ degrees and $\angle ADC$ is $29$ degrees. Since $OC$ is a radius and intersects $DA$ at $P$ , $\angle OPC$ is also $40$ degrees because it's the same angle as $\angle OCD$ . Now, triangle $\angle OCD$ $1$ is isosceles with two sides being radii of the circle, so $\angle OCD$ $2$ degrees.
Find Angle AOC: Since $_/OCP$ and $_/OPC$ are $40$ degrees each, the remaining angle $_/PCO$ in triangle OCP must be $100$ degrees $(180 - 40 - 40)$ because the sum of angles in a triangle is $180$ degrees.
Find Angle AOC: Since $\angle OCP$ and $\angle OPC$ are $40$ degrees each, the remaining angle $\angle PCO$ in triangle OCP must be $100$ degrees ( $180 - 40 - 40$ ) because the sum of angles in a triangle is $180$ degrees.Now, since $\angle ADC$ is $29$ degrees and $\angle PCO$ is $100$ degrees, $\angle OPC$ $1$ , which is the exterior angle for triangle OCP, is the sum of $\angle ADC$ and $\angle PCO$ . So, $\angle OPC$ $4$ degrees.
Find Angle AOC: Since $_/OCP$ and $_/OPC$ are $40$ degrees each, the remaining angle $_/PCO$ in triangle OCP must be $100$ degrees $(180 - 40 - 40)$ because the sum of angles in a triangle is $180$ degrees.Now, since $_/ADC$ is $29$ degrees and $_/PCO$ is $100$ degrees, $_/OPC$ $1$ , which is the exterior angle for triangle OCP, is the sum of $_/ADC$ and $_/PCO$ . So, $_/OPC$ $4$ degrees.b. To find $_/OPC$ $5$ , we look at triangle AOC. Since $_/OPC$ $1$ is $_/OPC$ $7$ degrees and it's an exterior angle to triangle AOC, $_/OPC$ $5$ must be $_/OPC$ $9$ degrees because the exterior angle is equal to the sum of the two opposite interior angles.