1. Problem 1: Find the distance from the switch to the lamp in a cubic room with side lengths 6 m. - The room is a cube with sides 6 m. - The lamp is at the center of the ceiling, so its coordinates are at the center of the top face: $(3, 3, 6)$. - The switch is at the center of one wall, for example, the wall at $x=0$, so its coordinates are $(0, 3, 3)$. - Distance between switch $(0,3,3)$ and lamp $(3,3,6)$ is: $$d = \sqrt{(3-0)^2 + (3-3)^2 + (6-3)^2} = \sqrt{3^2 + 0 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$$ 2. Problem 2: Find the area of the image of triangle $\triangle ABC$ under translation $T$ that maps $P(2,-1)$ to $P'(5,3)$. - Translation vector is $(a,b) = (5-2, 3-(-1)) = (3,4)$. - Translation preserves area. - Coordinates of $A(1,1)$, $B(4,2)$, $C(2,5)$. - Area of $\triangle ABC$ is: $$\text{Area} = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$ $$= \frac{1}{2} |1(2-5) + 4(5-1) + 2(1-2)| = \frac{1}{2} |-3 + 16 - 2| = \frac{1}{2} |11| = 5.5$$ - Area of image $\triangle A'B'C'$ is also $5.5$. 3. Problem 3: Given $90^\circ < \alpha < 180^\circ$ and $\cos \alpha = -\frac{4}{5}$, find $\sin \alpha$. - Since $90^\circ < \alpha < 180^\circ$, $\sin \alpha > 0$ and $\cos \alpha < 0$. - Use Pythagorean identity: $$\sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(-\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25}$$ - So, $$\sin \alpha = \sqrt{\frac{9}{25}} = \frac{3}{5}$$ Final answers: 1. Distance switch to lamp = $3\sqrt{2}$ m 2. Area of translated triangle = $5.5$ 3. $\sin \alpha = \frac{3}{5}$