Subject 3. Consider the adjacent angles $\Varangle A O B$ and $\Varangle B O C$, such that their bisectors [OM and $\left[O N\right.$ form an angle of $75^{\circ}$.

a) Determine $m(\Varangle A O B)$ and $m(\Varangle B O C)$ knowing that $3 \cdot m(\Varangle A O B)=2 \cdot m(\Varangle B O C)$.

b) If the ray [OT forms a right angle with the ray [OM such that points $M$ and $T$ are on the same side as point $B$ relative to point $A$. Calculate: $m(\Varangle T O N), m(\Varangle B O N)$ and $m(\Varangle B O T)$ and $m(\Varangle C O T)$.
## Solution:

a) Let $m(\angle A O B)=\alpha$ and $m(\angle B O C)=\beta$. We have:

$[O M]$ is the bisector of $\angle A O B \Rightarrow m(\angle A O M)=m(\angle M O B)=\frac{\alpha}{2}$;

$[O N]$ is the bisector of $\angle B O C \Rightarrow m(\angle B O N)=m(\angle N O C)=\frac{\beta}{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_9e90d71b69e1f57c0219g-2.jpg?height=334&width=654&top_left_y=1386&top_left_x=1272)

Since $m(\angle M O N)=75^{\circ}$, we have $\frac{\alpha}{2}+\frac{\beta}{2}=75^{\circ} \Leftrightarrow \alpha+\beta=150^{\circ} \Rightarrow \alpha=150^{\circ}-\beta$, (1).

But $3 \cdot m(\angle A O B)=2 \cdot m(\angle B O C) \Rightarrow 3 \alpha=2 \beta$, (2).

From (1) and (2) we have $3(150^{\circ}-\beta)=2 \beta \Leftrightarrow 450^{\circ}=5 \beta \Leftrightarrow \beta=90^{\circ}$ and $\alpha=150^{\circ}-90^{\circ}=60^{\circ}$.

Thus, $m(\angle A O B)=60^{\circ}$ and $m(\angle B O C)=90^{\circ}$.

b) We have $m(\angle M O T)=90^{\circ}$, from which we get: $m(\angle T O N)=90^{\circ}-[m(\angle N O B)+m(\angle M O B)]=15^{\circ}$;

$m(\angle B O N)=\frac{\beta}{2}=\frac{90^{\circ}}{2}=45^{\circ}$; $m(\angle B O T)=m(\angle B O N)+m(\angle T O N)=45^{\circ}+15^{\circ}=60^{\circ}$.

$m(\angle C O T)=m(\angle C O N)-m(\angle T O N)=45^{\circ}-15^{\circ}=30^{\circ}$.

| Details of the solution | Associated grading |
| :---: | :---: |
| Figure corresponding to the problem. | $1 \text{p}$ |
| a) Let $m(\angle A O B)=\alpha$ and $m(\angle B O C)=\beta$ and deduce that $\alpha+\beta=150^{\circ}$ | $1 \text{p}$ |
| Determine $m(\angle A O B)=60^{\circ}$ and $m(\angle B O C)=90^{\circ}$ | $1 \text{p}$ |
| b) Specify that $m(\angle M O T)=90^{\circ}$ and calculate $m(\angle T O N)=15^{\circ}$ | $1 \text{p}$ |
| $m(\angle B O N)=45^{\circ}$ | $1 \text{p}$ |
| $m(\angle B O T)=60^{\circ}$ | $1 \text{p}$ |
| $m(\angle C O T)=30^{\circ}$ | $1 \text{p}$ |