1. **Problem Statement:** Solve the simultaneous equations graphically: $$2x + 3y = 12$$ and $$x - y = 1$$. 2. **Formulas and Rules:** - To solve graphically, express each equation in terms of $$y$$: - From $$2x + 3y = 12$$, solve for $$y$$: $$3y = 12 - 2x \implies y = \frac{12 - 2x}{3}$$ - From $$x - y = 1$$, solve for $$y$$: $$y = x - 1$$ - The solution is the point where the two lines intersect. 3. **Intermediate Work:** - Set the two expressions for $$y$$ equal: $$\frac{12 - 2x}{3} = x - 1$$ - Multiply both sides by 3: $$12 - 2x = 3x - 3$$ - Rearrange: $$12 + 3 = 3x + 2x$$ $$15 = 5x$$ - Solve for $$x$$: $$x = 3$$ - Substitute $$x=3$$ into $$y = x - 1$$: $$y = 3 - 1 = 2$$ 4. **Answer:** The solution to the system is $$\boxed{(3, 2)}$$. 5. **Problem Statement:** Evaluate: $$\frac{2}{3} \csc^2 58^\circ - \frac{2}{3} \cot 58^\circ \tan 32^\circ - \frac{5}{3} \tan 13^\circ \tan 37^\circ \tan 45^\circ \tan 53^\circ \tan 77^\circ$$ 6. **Formulas and Rules:** - Use complementary angle identities: - $$\tan(90^\circ - \theta) = \cot \theta$$ - $$\csc^2 \theta = 1 + \cot^2 \theta$$ - $$\tan 45^\circ = 1$$ 7. **Intermediate Work:** - Note that $$\tan 32^\circ = \cot 58^\circ$$ since $$32^\circ + 58^\circ = 90^\circ$$ - So, $$\cot 58^\circ \tan 32^\circ = \cot 58^\circ \cdot \cot 58^\circ = \cot^2 58^\circ$$ - Also, $$\tan 13^\circ \tan 77^\circ = 1$$ (complementary angles) - $$\tan 37^\circ \tan 53^\circ = 1$$ (complementary angles) - So, $$\tan 13^\circ \tan 37^\circ \tan 45^\circ \tan 53^\circ \tan 77^\circ = (\tan 13^\circ \tan 77^\circ)(\tan 37^\circ \tan 53^\circ) \tan 45^\circ = 1 \times 1 \times 1 = 1$$ - Substitute back: $$\frac{2}{3} \csc^2 58^\circ - \frac{2}{3} \cot^2 58^\circ - \frac{5}{3} \times 1$$ - Use $$\csc^2 58^\circ = 1 + \cot^2 58^\circ$$: $$\frac{2}{3} (1 + \cot^2 58^\circ) - \frac{2}{3} \cot^2 58^\circ - \frac{5}{3}$$ - Simplify: $$\frac{2}{3} + \frac{2}{3} \cot^2 58^\circ - \frac{2}{3} \cot^2 58^\circ - \frac{5}{3} = \frac{2}{3} - \frac{5}{3} = -1$$ 8. **Answer:** The value is $$\boxed{-1}$$. 9. **Problem Statement:** Prove that if $$AD$$ and $$BC$$ are equal perpendiculars to a line segment $$AB$$, then $$CD$$ bisects $$AB$$. 10. **Proof:** - Given: $$AD$$ and $$BC$$ are perpendiculars from points $$A$$ and $$B$$ to line $$CD$$, and $$AD = BC$$. - Since $$AD$$ and $$BC$$ are equal and perpendicular to $$CD$$, triangles $$ADC$$ and $$BCD$$ are congruent by RHS (Right angle, Hypotenuse, Side) criterion. - Therefore, $$AC = BC$$. - Since $$C$$ lies on the line segment $$AB$$ and $$AC = BC$$, point $$C$$ is the midpoint of $$AB$$. - Hence, $$CD$$ bisects $$AB$$. 11. **Answer:** $$CD$$ bisects $$AB$$ as proved.