4. Consider the adjacent complementary angles $\square A O B$ and $\square B O C$, points $D, E \in \operatorname{Int}(\square A O B)$, such that $\square A O D \equiv \square D O E \equiv \square E O B$, and (OF is the bisector of angle $\square B O C$. Given that $\mathrm{m}(\square E O F)=39^{\circ}$, determine:

(4p) a) $\mathrm{m}(\square A O B)$ and $\mathrm{m}(\square B O C)$.

(3p) b) m( $\square D O M$ ), where ( $O M$ is the bisector of angle $\square E^{\prime} O C$, and ( $O E$ and ( $O E^{\prime}$ are opposite rays.

Adina Oancea

Note: All subjects are mandatory.

Effective working time: 2 hours.
4. a) $\mathrm{m}(\square E O B)+\mathrm{m}(\square B O F)=39^{\circ}$ ..... $(0.5 p)$
$\mathrm{m}(\square A O B)+\mathrm{m}(\square B O C)=90^{\circ}$ ..... $(0.5 p)$
$\square A O D \equiv \square D O E \equiv \square E O B$ and $\square B O F \equiv \square F O C \Rightarrow 3 \mathrm{~m}(\square E O B)+2 \mathrm{~m}(\square B O F)=90^{\circ}$ ..... (1p)
$\Rightarrow \mathrm{m}(\square E O B)=12^{\circ}$, thus $\Rightarrow \mathrm{m}(\square A O B)=3 \mathrm{~m}(\square E O B)=36^{\circ}$ ..... (1p)
$\Rightarrow \mathrm{m}(\square B O C)=54^{\circ}$ ..... (1p)
b) $\mathrm{m}\left(\square E O E^{\prime}\right)=180^{\circ} \Rightarrow \mathrm{m}\left(\square E^{\prime} O C\right)=180^{\circ}-\mathrm{m}(\square E O B)-\mathrm{m}(\square B O C)=114^{\circ}$ ..... (1p)
$\mathrm{m}\left(\square E^{\prime} O M\right)=\mathrm{m}(\square M O C)=57^{\circ}$ ..... (1p)
$\mathrm{m}(\square D O M)=2 \mathrm{~m}(\square E O B)+\mathrm{m}(\square B O C)+\mathrm{m}(\square C O M)=135^{\circ}$ ..... (1p)