1. Problem 1 statement: Given that $\sin\theta=5/13$ and $\theta$ is acute, find $\tan\theta$, $\csc\theta$, $\sin\theta$, $\cos\theta$, and $\cot\theta$. 2. Formula and rules: By definition, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$, $\csc\theta=\frac{1}{\sin\theta}$, and $\cot\theta=\frac{1}{\tan\theta}$. 3. Since $\theta$ is acute we take the triangle side lengths as positive. 4. From $\sin\theta=5/13$ we set opposite$=5$ and hypotenuse$=13$. 5. Compute the adjacent side by Pythagoras: $\text{adjacent}=\sqrt{13^2-5^2}=\sqrt{169-25}=\sqrt{144}=12$. 6. Compute $\cos\theta$: $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{12}{13}$. 7. Compute $\tan\theta$: $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{5}{12}$. 8. Compute $\csc\theta$: $\csc\theta=\frac{1}{\sin\theta}=\frac{1}{5/13}=\frac{13}{5}$. 9. Compute $\cot\theta$: $\cot\theta=\frac{1}{\tan\theta}=\frac{12}{5}$. 10. Final answers for Problem 1: $\tan\theta=\frac{5}{12}$. 11. Final answers for Problem 1: $\csc\theta=\frac{13}{5}$. 12. Final answers for Problem 1: $\sin\theta=\frac{5}{13}$. 13. Final answers for Problem 1: $\cos\theta=\frac{12}{13}$. 14. Final answers for Problem 1: $\cot\theta=\frac{12}{5}$. 15. Problem 2 statement: (a) Find the lengths of the lines joining the given pairs of points; (b) Find the gradients of the lines joining the given pairs of points; (c) The point M divides the line PQ where $P=(3,2)$ and $Q=(4,1)$ in the ratio $3:(-2)$, find the coordinates of the dividing point. 16. Formula for distance: For points $(x_1,y_1)$ and $(x_2,y_2)$, $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. 17. (a)(i) For points $(3,-5)$ and $(4,-3)$ compute differences: $\Delta x=4-3=1$ and $\Delta y=-3-(-5)=2$. 18. Distance: $d=\sqrt{1^2+2^2}=\sqrt{1+4}=\sqrt{5}$. 19. (a)(ii) For points $(-2,7)$ and $(3,-2)$ compute differences: $\Delta x=3-(-2)=5$ and $\Delta y=-2-7=-9$. 20. Distance: $d=\sqrt{5^2+(-9)^2}=\sqrt{25+81}=\sqrt{106}$. 21. Formula for gradient (slope): For points $(x_1,y_1)$ and $(x_2,y_2)$, $m=\frac{y_2-y_1}{x_2-x_1}$. 22. (b)(i) For points $(7,-9)$ and $(2,5)$ compute differences: $\Delta x=2-7=-5$ and $\Delta y=5-(-9)=14$. 23. Gradient: $m=\frac{14}{-5}=-\frac{14}{5}$. 24. (b)(ii) For points $(-2,-1)$ and $(6,7)$ compute differences: $\Delta x=6-(-2)=8$ and $\Delta y=7-(-1)=8$. 25. Gradient: $m=\frac{8}{8}=1$. 26. Section formula: If a point divides $P(x_1,y_1)$ and $Q(x_2,y_2)$ in the ratio $m:n$ then the coordinates are $\left(\frac{m x_2 + n x_1}{m+n},\frac{m y_2 + n y_1}{m+n}\right)$. 27. For part (c) take $P=(3,2)$, $Q=(4,1)$ and ratio $m:n=3:(-2)$ so $m=3$ and $n=-2$. 28. Compute denominator $m+n=3+(-2)=1$. 29. Compute x-coordinate: $x=\frac{3\cdot 4 + (-2)\cdot 3}{1}=\frac{12-6}{1}=6$. 30. Compute y-coordinate: $y=\frac{3\cdot 1 + (-2)\cdot 2}{1}=\frac{3-4}{1}=-1$. 31. So the dividing point M has coordinates $(6,-1)$. 32. Final answers for Problem 2: (a)(i) distance $=\sqrt{5}$. 33. Final answers for Problem 2: (a)(ii) distance $=\sqrt{106}$. 34. Final answers for Problem 2: (b)(i) gradient $=-\frac{14}{5}$. 35. Final answers for Problem 2: (b)(ii) gradient $=1$. 36. Final answers for Problem 2: (c) dividing point $= (6,-1)$.