### Instruction: Acute-angled $\triangle ABC$ is inscribed in a circle with center at $O$; $\stackrel \frown {AB} = 120^\circ$ and $\stackrel \frown {BC} = 72^\circ$.
A point $E$ is taken in minor arc $AC$ such that $OE$ is perpendicular to $AC$. Then the ratio of the magnitudes of $\angle OBE$ and $\angle BAC$ is:
$\textbf{(A)}\ \frac{5}{18}\qquad \textbf{(B)}\ \frac{2}{9}\qquad \textbf{(C)}\ \frac{1}{4}\qquad \textbf{(D)}\ \frac{1}{3}\qquad \textbf{(E)}\ \frac{4}{9}$.### Response: I know that the measure of an inscribed angle in a circle is half the measure of the central angle subtended by the same arc.
So, $\angle BAC = \frac{1}{2} \stackrel \frown {BC} = \frac{1}{2} \cdot 72^\circ = 36^\circ$.
I also know that the measure of an angle subtended by an arc at the center of the circle is equal to the measure of the arc itself.
So, $\angle AOB = \stackrel \frown {AB} = 120^\circ$.
Since $\triangle AOB$ is isosceles, with $OA = OB$, I can find the base angles by subtracting the vertex angle from $180^\circ$ and dividing by $2$.
So, $\angle OBA = \frac{1}{2}(180^\circ - 120^\circ) = 30^\circ$.
Now, I can use the fact that $OE$ is perpendicular to $AC$ to find $\angle OEA$.
Since $\triangle OAE$ is right, and $\angle AOE = \frac{1}{2} \stackrel \frown {AE}$, I can use the Pythagorean theorem to find $AE$.
I have $OE^2 + EA^2 = OA^2$, and $OA$ is the radius of the circle, which I can call $r$.
Substituting, I get $r^2 - OE^2 = EA^2$.
Since $OE$ is perpendicular to $AC$, it is also the altitude of $\triangle AOB$, and I can use the area formula to find it.
I have $\frac{1}{2} AB \cdot OE = [\triangle AOB]$, and $[\triangle AOB] = \frac{1}{2} r^2 \sin \angle AOB = \frac{1}{2} r^2 \sin 120^\circ = \frac{r^2 \sqrt{3}}{4}$.
Simplifying, I get $AB \cdot OE = \frac{r^2 \sqrt{3}}{2}$, and $AB = 2r \sin \angle OBA = 2r \sin 30^\circ = r$.
Therefore, $OE = \frac{r \sqrt{3}}{2}$.
Plugging this back into the Pythagorean equation, I get $r^2 - (\frac{r \sqrt{3}}{2})^2 = EA^2$.
Simplifying, I get $\frac{r^2}{4} = EA^2$, and $EA = \frac{r}{2}$.
Now, I can use the trigonometric ratios to find $\angle OEA$.
I have $\tan \angle OEA = \frac{OE}{EA} = \frac{r \sqrt{3}}{2} \div \frac{r}{2} = \sqrt{3}$.
So, $\angle OEA = \tan^{-1} \sqrt{3} = 60^\circ$.
Finally, I can use the angle sum formula for $\triangle OBE$ to find $\angle OBE$.
I have $\angle OBE = 180^\circ - \angle OBA - \angle OEA = 180^\circ - 30^\circ - 60^\circ = 90^\circ$.
Therefore, the ratio of the magnitudes of $\angle OBE$ and $\angle BAC$ is $\frac{90^\circ}{36^\circ} = \frac{5}{2}$.