1. **Problem statement:** Given a rectangular prism with side lengths $a=11.6$ cm, $b=9.7$ cm, and $c=14.3$ cm, and $M$ as the midpoint of segment $SR$, calculate the size of angle $WWMT$ to 1 decimal place. 2. **Understanding the problem:** We need to find the angle formed at point $W$ by points $W$, $M$, and $T$. Since $M$ is the midpoint of $SR$, we first find coordinates of points $S$, $R$, and $M$ assuming a coordinate system. 3. **Assign coordinates:** - Let $W$ be at origin $(0,0,0)$. - Since $a=11.6$ cm, $b=9.7$ cm, and $c=14.3$ cm correspond to edges along $x$, $y$, and $z$ axes respectively, we can assign: - $S = (0,b,0) = (0,9.7,0)$ - $R = (a,b,0) = (11.6,9.7,0)$ - $T = (a,0,c) = (11.6,0,14.3)$ 4. **Find midpoint $M$ of $SR$:** $$M = \left(\frac{0+11.6}{2}, \frac{9.7+9.7}{2}, \frac{0+0}{2}\right) = (5.8, 9.7, 0)$$ 5. **Vectors for angle calculation:** - Vector $\overrightarrow{WM} = M - W = (5.8, 9.7, 0)$ - Vector $\overrightarrow{WT} = T - W = (11.6, 0, 14.3)$ 6. **Calculate dot product:** $$\overrightarrow{WM} \cdot \overrightarrow{WT} = (5.8)(11.6) + (9.7)(0) + (0)(14.3) = 67.28$$ 7. **Calculate magnitudes:** $$|\overrightarrow{WM}| = \sqrt{5.8^2 + 9.7^2 + 0^2} = \sqrt{33.64 + 94.09} = \sqrt{127.73} \approx 11.3$$ $$|\overrightarrow{WT}| = \sqrt{11.6^2 + 0^2 + 14.3^2} = \sqrt{134.56 + 204.49} = \sqrt{339.05} \approx 18.41$$ 8. **Calculate angle $\theta$ using dot product formula:** $$\cos \theta = \frac{\overrightarrow{WM} \cdot \overrightarrow{WT}}{|\overrightarrow{WM}| |\overrightarrow{WT}|} = \frac{67.28}{11.3 \times 18.41} = \frac{67.28}{208.03} \approx 0.3233$$ 9. **Find angle in degrees:** $$\theta = \cos^{-1}(0.3233) \approx 71.2^\circ$$ **Final answer:** The size of angle $WWMT$ is approximately $71.2^\circ$ to 1 decimal place.