1. **Problem Statement:** Calculate the angle between line segment $AP$ and the rectangular base $ABCD$ of the triangular prism. 2. **Given Data:** - $AB = 12$ cm - $BC = 6$ cm - $PC = 4$ cm - $ riangle BCP$ is right-angled at $C$ (angle $BCP = 90^\circ$) - $QDC$ is right-angled at $D$ 3. **Understanding the problem:** The base $ABCD$ is a rectangle lying in a plane. We want the angle between the line $AP$ and this base plane. 4. **Approach:** The angle between a line and a plane is complementary to the angle between the line and the plane's normal vector. So, $$\theta = 90^\circ - \alpha$$ where $\alpha$ is the angle between $AP$ and the normal vector to the base $ABCD$. 5. **Step 1: Find coordinates for points assuming $A$ at origin:** - Let $A = (0,0,0)$ - Since $AB = 12$ cm along x-axis, $B = (12,0,0)$ - $BC = 6$ cm along y-axis, so $C = (12,6,0)$ - $D$ is above $A$ and $C$ with height 4 cm, so $D = (0,6,0)$ - $P$ is above $C$ by 4 cm, so $P = (12,6,4)$ 6. **Step 2: Vector $AP$:** $$\vec{AP} = P - A = (12,6,4) - (0,0,0) = (12,6,4)$$ 7. **Step 3: Normal vector to base $ABCD$:** Since $ABCD$ lies in the $xy$-plane, its normal vector is along the $z$-axis: $$\vec{n} = (0,0,1)$$ 8. **Step 4: Calculate angle $\alpha$ between $\vec{AP}$ and $\vec{n}$ using dot product:** $$\cos \alpha = \frac{\vec{AP} \cdot \vec{n}}{|\vec{AP}||\vec{n}|} = \frac{4}{\sqrt{12^2 + 6^2 + 4^2} \times 1} = \frac{4}{\sqrt{144 + 36 + 16}} = \frac{4}{\sqrt{196}} = \frac{4}{14} = \frac{2}{7}$$ 9. **Step 5: Calculate $\alpha$:** $$\alpha = \cos^{-1}\left(\frac{2}{7}\right) \approx 73.4^\circ$$ 10. **Step 6: Calculate angle $\theta$ between $AP$ and base $ABCD$:** $$\theta = 90^\circ - 73.4^\circ = 16.6^\circ$$ **Final answer:** The angle between $AP$ and the rectangular base $ABCD$ is approximately $16.6^\circ$.