1. **Problem statement:** We have an isosceles triangle ABC with AB = AC = 13 cm and BC = 10 cm lying in a horizontal plane. R is the midpoint of BC, Q is the midpoint of AR, and a stick PQ of length $6\sqrt{3}$ cm stands vertically at Q. We need to find: (a) the angle between the planes PBC and ABC. (b) the angle $\angle PBQ$. 2. **Step 1: Find coordinates of points A, B, C, R, Q, and P** Place triangle ABC in the xy-plane for convenience. Let B = (0,0,0) and C = (10,0,0) since BC = 10 cm. Since ABC is isosceles with AB = AC = 13 cm, point A lies on the perpendicular bisector of BC. The midpoint R of BC is at (5,0,0). 3. **Find A's coordinates:** Let A = (5,y,0). Distance AB = 13 cm: $$AB = \sqrt{(5-0)^2 + (y-0)^2} = 13$$ $$\Rightarrow \sqrt{25 + y^2} = 13$$ $$\Rightarrow 25 + y^2 = 169$$ $$\Rightarrow y^2 = 144$$ $$\Rightarrow y = 12$$ (taking positive since A is above BC line) So, A = (5,12,0). 4. **Find Q, midpoint of AR:** Coordinates of R = (5,0,0), A = (5,12,0) $$Q = \left(\frac{5+5}{2}, \frac{12+0}{2}, \frac{0+0}{2}\right) = (5,6,0)$$ 5. **Find P, vertical from Q with length $6\sqrt{3}$:** Since PQ is vertical, P = (5,6,$6\sqrt{3}$). 6. **Find vectors in plane PBC:** Vector PB = B - P = (0-5, 0-6, 0 - 6\sqrt{3}) = (-5, -6, -6\sqrt{3}) Vector PC = C - P = (10-5, 0-6, 0 - 6\sqrt{3}) = (5, -6, -6\sqrt{3}) 7. **Find normal vector to plane PBC:** $$\vec{n}_{PBC} = \vec{PB} \times \vec{PC}$$ Calculate cross product: $$\vec{n}_{PBC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & -6 & -6\sqrt{3} \\ 5 & -6 & -6\sqrt{3} \end{vmatrix}$$ $$= \mathbf{i}((-6)(-6\sqrt{3}) - (-6)(-6\sqrt{3})) - \mathbf{j}((-5)(-6\sqrt{3}) - 5(-6\sqrt{3})) + \mathbf{k}((-5)(-6) - 5(-6))$$ Calculate each component: - $\mathbf{i}$: $36\sqrt{3} - 36\sqrt{3} = 0$ - $\mathbf{j}$: $30\sqrt{3} - (-30\sqrt{3}) = 30\sqrt{3} + 30\sqrt{3} = 60\sqrt{3}$ - $\mathbf{k}$: $30 - (-30) = 60$ So, $$\vec{n}_{PBC} = (0, -60\sqrt{3}, 60)$$ (note the minus sign for j component) 8. **Find normal vector to plane ABC:** Since ABC lies in the horizontal plane z=0, its normal vector is $$\vec{n}_{ABC} = (0,0,1)$$ 9. **Find angle between planes PBC and ABC:** Angle between planes equals angle between their normal vectors. Use dot product formula: $$\cos \theta = \frac{\vec{n}_{PBC} \cdot \vec{n}_{ABC}}{|\vec{n}_{PBC}||\vec{n}_{ABC}|}$$ Calculate dot product: $$\vec{n}_{PBC} \cdot \vec{n}_{ABC} = 0*0 + (-60\sqrt{3})*0 + 60*1 = 60$$ Calculate magnitudes: $$|\vec{n}_{PBC}| = \sqrt{0^2 + (-60\sqrt{3})^2 + 60^2} = \sqrt{0 + 3600*3 + 3600} = \sqrt{10800 + 3600} = \sqrt{14400} = 120$$ $$|\vec{n}_{ABC}| = 1$$ So, $$\cos \theta = \frac{60}{120*1} = \frac{1}{2}$$ $$\Rightarrow \theta = \cos^{-1}(\frac{1}{2}) = 60^\circ$$ 10. **Find angle $\angle PBQ$:** Points: P = (5,6,6\sqrt{3}), B = (0,0,0), Q = (5,6,0) Vectors: $$\vec{PB} = B - P = (-5, -6, -6\sqrt{3})$$ $$\vec{QB} = B - Q = (-5, -6, 0)$$ Angle between $\vec{PB}$ and $\vec{QB}$ is $\angle PBQ$. Use dot product: $$\cos \alpha = \frac{\vec{PB} \cdot \vec{QB}}{|\vec{PB}||\vec{QB}|}$$ Calculate dot product: $$\vec{PB} \cdot \vec{QB} = (-5)(-5) + (-6)(-6) + (-6\sqrt{3})(0) = 25 + 36 + 0 = 61$$ Calculate magnitudes: $$|\vec{PB}| = \sqrt{(-5)^2 + (-6)^2 + (-6\sqrt{3})^2} = \sqrt{25 + 36 + 36*3} = \sqrt{25 + 36 + 108} = \sqrt{169} = 13$$ $$|\vec{QB}| = \sqrt{(-5)^2 + (-6)^2 + 0^2} = \sqrt{25 + 36} = \sqrt{61}$$ So, $$\cos \alpha = \frac{61}{13 \times \sqrt{61}} = \frac{61}{13\sqrt{61}} = \frac{\sqrt{61}}{13}$$ Calculate angle: $$\alpha = \cos^{-1}\left(\frac{\sqrt{61}}{13}\right) \approx 53.07^\circ$$ **Final answers:** (a) Angle between planes PBC and ABC is $60^\circ$. (b) Angle $\angle PBQ$ is approximately $53.07^\circ$.