4. An obtuse angle $\Varangle A O B$ and points $C, D$ located in opposite half-planes relative to the line $A B$ are given, such that $\mathrm{m}(\Varangle C O D)=80^{\circ}$.

(4p) a) If $[O N$ is the bisector of $\Varangle A O C$, $[O M$ is the bisector of $\Varangle B O D$, and $\mathrm{m}(\Varangle B O C)=$ $140^{\circ} 15'$ $30''$, calculate the measure of $\Varangle M O N$.

(3p) b) If [OE is the ray opposite to the ray $[O D$, calculate the measure of $\Varangle B O E$.

SGM12/2013

Note: All questions are mandatory.

Actual working time: 2 hours.

## Grading Rubric OLM Grade 6, 2014
4. a) Fig. ..... (1p)
$\mathrm{m}(\angle M O N)=(\mathrm{m}(\angle M O D)+\mathrm{m}(\angle D O A)+\mathrm{m}(\angle A O N)=$

$=\mathrm{m}(\angle B O D): 2+\mathrm{m}(\angle D O A)+\mathrm{m}(\angle A O C): 2=$

$=(\mathrm{m}(\angle B O D)+2 \mathrm{~m}(\angle D O A)+\mathrm{m}(\angle A O C)): 2=$ $=(\mathrm{m}(\angle B O A)+\mathrm{m}(\angle D O C)): 2=$

$=(180^{\circ}+80^{\circ}): 2=130^{\circ}$

(3p)
b) $\mathrm{m}(\angle B O D)=360^{\circ}-[\mathrm{m}(\angle C O D)+\mathrm{m}(\angle B O C)]=$
$=360^{\circ}-\left[80^{\circ}+140^{\circ} 15^{\prime} 30^{\prime \prime}\right]=$

![](https://cdn.mathpix.com/cropped/2024_06_07_a96920e3230ef7108286g-2.jpg?height=417&width=731&top_left_y=2279&top_left_x=1162)

$$
\begin{aligned}
& =139^{\circ} 44^{\prime} 30^{\prime \prime} \\
& \text { of } \\
& \mathrm{m}(\angle B O E)=\mathrm{m}(\angle D O E)-\mathrm{m}(\angle B O D)= \\
& \left.=180^{\circ}-139^{\circ} 44^{\prime} 30^{\prime \prime}=40^{\circ} 15^{\prime} 30^{\prime \prime} \quad \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . . . . . . .1 p\right)
\end{aligned}
$$