4. Triangle $ABC$ has side $[BC]$ included in a plane $\alpha$. Knowing that $AB=a$, $BC=2a$, $\mathrm{m}(\widehat{ABC})=30^{\circ}$, and the sine of the dihedral angle between the planes $(ABC)$ and $\alpha$ is equal to $\frac{2 \sqrt{2}}{3}$, determine:

a) The length of the segment $\left[A A^{\prime}\right]$, where $A^{\prime}$ is the projection of $A$ onto the plane $\alpha$.

b) The distance from $A^{\prime}$ to the plane $(ABC)$.

## Grading Scale for 8th Grade
4. a) Let $D$ be the projection of $A$ onto $BC$.

In triangle $ABD$, $AD = AB \sin 30^\circ$, so $AD = \frac{a}{2}$.

In triangle $ADA'$, $\sin \left(\widehat{ADA'}\right) = \frac{2\sqrt{2}}{3} = \frac{AA'}{\frac{a}{2}}$ and $AA' = \frac{a\sqrt{2}}{3}$.

b) In the right triangle $ADA'$, $AD = \frac{a}{2}$, $AA' = \frac{a\sqrt{2}}{3}$, and it follows that $A'D = \frac{a}{6}$...

Drawing $A'E \perp AD$, $E \in (AD)$, it follows that $A'E \perp (ABC)$, because $A'E$ is perpendicular to $BC \left(BC \perp (ADA')\right)$.

Thus, $A'E = d(A', (ABC))$, and in the right triangle $ADA'$, $A'E = \frac{AA' \cdot A'D}{AD} = \frac{\frac{a\sqrt{2}}{3} \cdot \frac{a}{6}}{\frac{a}{2}} = \frac{a\sqrt{2}}{9}$

(in the figure, $u$ is the measure of the dihedral angle).

![](https://cdn.mathpix.com/cropped/2024_06_07_d22b3280a5e56a183b72g-3.jpg?height=1022&width=1345&top_left_y=991&top_left_x=193)

Note: Any correct solution different from the one in the grading rubric is awarded the maximum score.