## Subject III.

Angles $\Varangle A O B$ and $\Varangle A O C$ are supplementary, and their bisectors (OM and (ON form an angle with a measure equal to $60^{\circ}$. Calculate the measures of the two angles.
## Subject III

Angles $\Varangle A O B$ and $\Varangle A O C$ are supplementary, and their bisectors (OM and (ON form an angle with a measure equal to $60^{\circ}$. Calculate the measures of the two angles.

## Solution

If $\left(O B\right.$ and $\left(O C\right.$ were located on opposite sides of side $\left(O A\right.$, then $\mathrm{m}(\Varangle B O C)=180^{\circ}$, so $\mathrm{m}(\Varangle M O N)=90^{\circ}$, which contradicts the statement (1); let $\left(O B \subset \operatorname{Int}(\Varangle A O C) ; \mathrm{m}(\Varangle A O M)=\mathrm{m}(\Varangle M O B)=\frac{1}{2} \cdot \mathrm{m}(\Varangle A O B)=x\right.$, and $\mathrm{m}(\Varangle A O N)$ $=\mathrm{m}(\Varangle N O C)=\frac{1}{2} \cdot \mathrm{m}(\Varangle A O C)=y(2) ;$ from (2) and the statement, it follows that

$2 x+2 y=180^{\circ} \Leftrightarrow x+y=90^{\circ}(3) ; \mathrm{m}(\Varangle M O N)=\mathrm{m}(\Varangle A O C)-\mathrm{m}(\Varangle N O C)-\mathrm{m}(\Varangle A O M) \Rightarrow$

$2 y-y-x=60^{\circ} \Leftrightarrow y-x=60^{\circ}$ (4); adding (3) and (4) member by member, we get $y=$ $=75^{\circ}$, so $x=15^{\circ}, \mathrm{m}(\Varangle A O B)=30^{\circ}$ and $\mathrm{m}(\Varangle A O C)=150^{\circ}$.

## Grading Rubric:

```
- analyzes (1) ............ (1p)
- writes (2) ................... (1p)
- deduces (3) .................. (1p)
- deduces (4) .................. (2p)
- finalizes ................... (2p)
Total \(7 \mathbf{p}\)
```