1. **Problem 7:** Given triangles $\triangle ABC$ and $\triangle DEF$ with right angles at $\angle C$ and $\angle F$, and $\sin A = \cos D$, determine the relationship between pairs of angles. - Since $\sin A = \cos D$, and for acute angles $\sin \theta = \cos(90^\circ - \theta)$, it follows that $A$ and $D$ are complementary: $A + D = 90^\circ$. - Both triangles have right angles at $C$ and $F$, so $\angle C \cong \angle F = 90^\circ$. - In right triangles, the sum of the other two angles is $90^\circ$, so $B = 90^\circ - A$ and $E = 90^\circ - D$. Now, analyze each pair: - $\angle A$ and $\angle D$: Complementary (since $A + D = 90^\circ$) - $\angle B$ and $\angle F$: $B + F = (90^\circ - A) + 90^\circ = 180^\circ - A$, not complementary or congruent, so Neither - $\angle B$ and $\angle E$: Since $B = 90^\circ - A$ and $E = 90^\circ - D$, and $A + D = 90^\circ$, then $B = E$, so Congruent - $\angle A$ and $\angle E$: $A + E = A + (90^\circ - D) = A + 90^\circ - D = 90^\circ + (A - D)$, not complementary or congruent, so Neither - $\angle E$ and $\angle D$: $E + D = (90^\circ - D) + D = 90^\circ$, Complementary - $\angle B$ and $\angle D$: $B + D = (90^\circ - A) + D = 90^\circ + (D - A)$, Neither 2. **Problem 8:** Find $\sin(90^\circ - 40^\circ)$. - Use the complementary angle identity: $\sin(90^\circ - \theta) = \cos \theta$. - So, $\sin(90^\circ - 40^\circ) = \cos 40^\circ$. - From the triangle, $\cos 40^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{BC}$. Answer: D) $\cos 40^\circ$, or $\frac{AB}{BC}$. 3. **Problem 9:** If $\sin X = \cos Y$ for acute angles $X$ and $Y$, how are the angles related? - Since $\sin X = \cos Y$ and $\sin \theta = \cos(90^\circ - \theta)$, it follows $X = 90^\circ - Y$. - Therefore, $X$ and $Y$ are complementary. Answer: B) complementary. 4. **Problem 10:** Find $\cos(90^\circ - 25^\circ)$. - Using the complementary angle identity: $\cos(90^\circ - \theta) = \sin \theta$. - So, $\cos(90^\circ - 25^\circ) = \sin 25^\circ$. - From the triangle, $\sin 25^\circ = \frac{EF}{DF}$. Answer: A) $\sin 65^\circ$, or $\frac{EF}{DF}$ (Note: $\sin 65^\circ = \sin(90^\circ - 25^\circ) = \cos 25^\circ$, so the correct choice is A). 5. **Problem 11:** Find the cosine ratio of $\angle Z$ in $\triangle XYZ$ with right angle at $Y$, $XY = 6$, $YZ = 4$. - First, find hypotenuse $XZ$ using Pythagoras: $$XZ = \sqrt{XY^2 + YZ^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}.$$ - Cosine of $\angle Z$ is adjacent over hypotenuse. Adjacent side to $Z$ is $YZ = 4$. - So, $$\cos Z = \frac{YZ}{XZ} = \frac{4}{2\sqrt{13}} = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13} \quad \text{(rationalized)}.$$ Answer: B) $\frac{2\sqrt{13}}{13}$.