Question: 336 EXERCISE 15F.3 1 A cube has sides of length 10 cm. Find the angle between the plane defined by BGD and plane ABCD. 2 4 cm 10 cm 10 cm A P S D Q R B C Find the angle that plane ASC makes with plane ABCD. 3 The slant edges of the pyramid opposite are 5 cm long. Find the angle between planes BCE and ABCD. 3 cm 5 cm A D N B E C 4 The pyramid of Cheops in Egypt has a height of 145 m and has base lengths 230 m. Find the angle between a sloping face and the base. position_hint: top-right, center, bottom-left

1. **Problem:** Find the angle between the plane defined by points $B$, $G$, $D$ and the plane $ABCD$ of a cube with side length $10$ cm. 2. **Understanding the problem:** The cube has side length $10$ cm. Plane $ABCD$ is the base of the cube (a square). Points $B$ and $D$ lie on the base, and $G$ is a vertex on the top face diagonally opposite $B$ and $D$. 3. **Step 1: Define coordinates for points** - Let $A = (0,0,0)$, $B = (10,0,0)$, $C = (10,10,0)$, $D = (0,10,0)$ (base square in $xy$-plane). - The cube height is $10$, so the top face points are $E = (0,0,10)$, $F = (10,0,10)$, $G = (10,10,10)$, $H = (0,10,10)$. 4. **Step 2: Find vectors in plane $BGD$** - Vector $\overrightarrow{BG} = G - B = (10,10,10) - (10,0,0) = (0,10,10)$ - Vector $\overrightarrow{BD} = D - B = (0,10,0) - (10,0,0) = (-10,10,0)$ 5. **Step 3: Find normal vector to plane $BGD$** - Normal vector $\mathbf{n_1} = \overrightarrow{BG} \times \overrightarrow{BD}$ Calculate cross product: $$\mathbf{n_1} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 10 & 10 \\ -10 & 10 & 0 \end{vmatrix} = \mathbf{i}(10 \cdot 0 - 10 \cdot 10) - \mathbf{j}(0 \cdot 0 - (-10) \cdot 10) + \mathbf{k}(0 \cdot 10 - (-10) \cdot 10)$$ $$= \mathbf{i}(0 - 100) - \mathbf{j}(0 + 100) + \mathbf{k}(0 + 100) = (-100, -100, 100)$$ 6. **Step 4: Find normal vector to plane $ABCD$** - Plane $ABCD$ lies in the $xy$-plane, so its normal vector is $\mathbf{n_2} = (0,0,1)$. 7. **Step 5: Find angle $\theta$ between planes using their normal vectors** - Formula: $$\cos \theta = \frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{||\mathbf{n_1}|| \cdot ||\mathbf{n_2}||}$$ Calculate dot product: $$\mathbf{n_1} \cdot \mathbf{n_2} = (-100)(0) + (-100)(0) + (100)(1) = 100$$ Calculate magnitudes: $$||\mathbf{n_1}|| = \sqrt{(-100)^2 + (-100)^2 + 100^2} = \sqrt{10000 + 10000 + 10000} = \sqrt{30000} = 100\sqrt{3}$$ $$||\mathbf{n_2}|| = 1$$ Calculate cosine: $$\cos \theta = \frac{100}{100\sqrt{3} \times 1} = \frac{1}{\sqrt{3}}$$ 8. **Step 6: Calculate angle** $$\theta = \arccos\left(\frac{1}{\sqrt{3}}\right) = 54.7356^\circ$$ **Final answer:** The angle between plane $BGD$ and plane $ABCD$ is approximately $54.74^\circ$.