3. On the plane of rectangle $\mathrm{ABCD}$, perpendicular $\mathrm{PD}$ is raised, with $\mathrm{P} \notin(\mathrm{ABC})$.

If $\mathrm{AM} \perp \mathrm{PB}, \mathrm{CN} \perp \mathrm{PB}(\mathrm{M}, \mathrm{N} \in \mathrm{PB}), \mathrm{MN}=3 \mathrm{~cm}, \mathrm{AB}=8 \mathrm{~cm}, \mathrm{BC}=4 \mathrm{~cm}$, calculate:

a) the length of segment (PB);

b) the tangent of the measure of the dihedral angle between planes (ABC) and (APC).

Prof. Petru Nicuţă, Rădăuţi
## Solution:

a) - If $\mathrm{PD} \perp(\mathrm{ABC})$ and $\mathrm{DC} \perp \mathrm{BC} \Rightarrow(\mathrm{T} .3 \perp) \mathrm{PC} \perp \mathrm{BC}$; similarly $\mathrm{PA} \perp \mathrm{AB}$.

Applying the cathetus theorem: - in $\triangle \mathrm{PBC}$ we get: $\mathrm{BC}^{2}=\mathrm{BN} \cdot \mathrm{PB}$

- in $\triangle \mathrm{PAB}$ we get: $\mathrm{AB}^{2}=\mathrm{BM} \cdot \mathrm{PB}$

From the difference of the relations we have: $\mathrm{AB}^{2}-\mathrm{BC}^{2}=\mathrm{PB} \cdot(\mathrm{BM}-\mathrm{BN}) \Rightarrow 48=3 \cdot \mathrm{PB}$, from which $\mathrm{PB}=16 \mathrm{~cm}$.

b) - If $\mathrm{DE} \perp \mathrm{AC} \Rightarrow(\mathrm{T} .3 \perp) \mathrm{PE} \perp \mathrm{AC} \Rightarrow \Varangle[(\mathrm{ABC}),(\mathrm{APC})]=\Varangle(\mathrm{DE}, \mathrm{PE})=\Varangle \mathrm{DEP}$

- in $\triangle \mathrm{DEP}$ right-angled at $\mathrm{D}: \operatorname{tg} \Varangle \mathrm{DEP}=\frac{P D}{D E}$
- Applying the Pythagorean theorem: - in $\triangle \mathrm{ABC}$ : $\mathrm{AC}=4 \sqrt{5} \mathrm{~cm}$;

$$
\text { - in } \triangle \mathrm{PDB}: \mathrm{PD}=4 \sqrt{11} \mathrm{~cm}
$$

- In $\triangle \mathrm{DAC}: \mathrm{DE} \perp \mathrm{AC} \Rightarrow \mathrm{DE}=\frac{D A \cdot D C}{A C}=\frac{8 \sqrt{5}}{5} \mathrm{~cm}$, thus $\operatorname{tg} \Varangle \mathrm{DEP}=\frac{\sqrt{55}}{2}$


## Grading:

| a) $\mathrm{PC} \perp \mathrm{BC}$ and $\mathrm{PA} \perp \mathrm{AB}$ | $1 \mathrm{p}$ |
| :---: | :---: |
| $\mathrm{BC}^{2}=\mathrm{BN} \cdot \mathrm{PB}$ and $\mathrm{AB}^{2}=\mathrm{BM} \cdot \mathrm{PB}$........................................................................................... | $1 \mathrm{p}$ |
| ![](https://cdn.mathpix.com/cropped/2024_06_07_8506ae9367b2ce5a4f31g-2.jpg?height=53&width=1591&top_left_y=2337&top_left_x=113) | $1 \mathrm{p}$ |
| b) $\Varangle[(\mathrm{ABC}),(\mathrm{APC})]=\Varangle(\mathrm{DE}, \mathrm{PE})=\Varangle \mathrm{DEP}$ | $1 \mathrm{p}$ |
| ![](https://cdn.mathpix.com/cropped/2024_06_07_8506ae9367b2ce5a4f31g-2.jpg?height=66&width=1591&top_left_y=2436&top_left_x=113) | $1 \mathrm{p}$ |

![](https://cdn.mathpix.com/cropped/2024_06_07_8506ae9367b2ce5a4f31g-3.jpg?height=306&width=1751&top_left_y=115&top_left_x=98)