Problem 2. The rays $[O A,[O B,[O C$ and $[O D$ form three angles that do not have common interiors, $[O X$ is the bisector of angle $A O B$, and $[O Y$ is the bisector of angle $C O D$. If $[O X \perp[O Y$, the measure of angle $A O B$ is three times greater than the measure of angle $B O C$, and the measure of angle $C O D$ is equal to the measure of angle $B O C$. Determine the measure of angle $B O C$ and the measure of angle $A O D$.
## Solution. 1p by default

![](https://cdn.mathpix.com/cropped/2024_06_07_a95b52d6620e9e007274g-2.jpg?height=663&width=1648&top_left_y=1870&top_left_x=204)

| $\left.\begin{array}{l}\text { If } \mathrm{m} \angle(B O C)=\mathrm{c}, \text { then } \\ 2 a=3 c \\ 2 b=c\end{array}\right\}$. But $a+b+c=90^{\circ} \Rightarrow \frac{3 c}{2}+c+\frac{c}{2}=90^{\circ}$ | 2p |
| :--- | :--- |
| From which $3 c=90^{\circ} \Leftrightarrow c=30^{\circ}$. | $1 p$ |
| Therefore $\mathrm{m} \angle(B O C)=30^{\circ}$ and |  |
| $\mathrm{m} \angle(A O D)=\mathrm{m} \angle(A O B)+\mathrm{m} \angle(B O C)+\mathrm{m} \angle(C O D)=$ |  |
| $=90^{\circ}+30^{\circ}+30^{\circ}=150^{\circ}$. |  |