1. Given that $\sin \theta = \frac{5}{13}$ and $\theta$ is acute, find: (a) $\tan \theta$ (b) $\csc \theta$ (c) $\sin \theta$ (d) $\cos \theta$ (e) $\cot \theta$ 2. (a) Find the lengths of the lines joining the following pairs of points: i. $(3, -5)$ and $(4, -3)$ ii. $(-2, 7)$ and $(3, -2)$ (b) Find the gradients of the lines joining the following pairs of points: i. $(7, -9)$ and $(2, 5)$ ii. $(-2, -1)$ and $(6, 7)$ (c) The point $M$ divides the line $PQ$, where the coordinates of $P(3,2)$ and $Q(4,1)$ respectively in the ratio $3: -2$. Find the coordinates of $M$. --- ### Step 1: Trigonometric values 1. We know $\sin \theta = \frac{5}{13}$ and $\theta$ is acute. 2. Use the Pythagorean identity: $$\sin^2 \theta + \cos^2 \theta = 1$$ 3. Substitute $\sin \theta$: $$\left(\frac{5}{13}\right)^2 + \cos^2 \theta = 1$$ 4. Calculate: $$\frac{25}{169} + \cos^2 \theta = 1$$ 5. Solve for $\cos^2 \theta$: $$\cos^2 \theta = 1 - \frac{25}{169} = \frac{169}{169} - \frac{25}{169} = \frac{144}{169}$$ 6. Since $\theta$ is acute, $\cos \theta = \frac{12}{13}$. 7. Calculate $\tan \theta$: $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$$ 8. Calculate $\csc \theta$: $$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{5}{13}} = \frac{13}{5}$$ 9. $\sin \theta$ is given as $\frac{5}{13}$. 10. $\cos \theta$ is $\frac{12}{13}$. 11. Calculate $\cot \theta$: $$\cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{5}{12}} = \frac{12}{5}$$ ### Step 2: Lengths of lines Length formula between points $(x_1,y_1)$ and $(x_2,y_2)$: $$\text{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ (a) i. Between $(3, -5)$ and $(4, -3)$: $$\sqrt{(4-3)^2 + (-3+5)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$$ (a) ii. Between $(-2, 7)$ and $(3, -2)$: $$\sqrt{(3+2)^2 + (-2-7)^2} = \sqrt{5^2 + (-9)^2} = \sqrt{25 + 81} = \sqrt{106}$$ ### Step 3: Gradients of lines Gradient formula: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ (b) i. Between $(7, -9)$ and $(2, 5)$: $$m = \frac{5 - (-9)}{2 - 7} = \frac{14}{-5} = -\frac{14}{5}$$ (b) ii. Between $(-2, -1)$ and $(6, 7)$: $$m = \frac{7 - (-1)}{6 - (-2)} = \frac{8}{8} = 1$$ ### Step 4: Coordinates of point $M$ Point $M$ divides $PQ$ in ratio $3 : -2$. Coordinates formula for division in ratio $m:n$: $$M = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$$ Here, $P(3,2)$, $Q(4,1)$, $m=3$, $n=-2$. Calculate denominator: $$m + n = 3 + (-2) = 1$$ Calculate $x$ coordinate: $$x_M = \frac{3 \times 4 + (-2) \times 3}{1} = \frac{12 - 6}{1} = 6$$ Calculate $y$ coordinate: $$y_M = \frac{3 \times 1 + (-2) \times 2}{1} = \frac{3 - 4}{1} = -1$$ So, $M = (6, -1)$. --- ### Final answers: 1. (a) $\tan \theta = \frac{5}{12}$ (b) $\csc \theta = \frac{13}{5}$ (c) $\sin \theta = \frac{5}{13}$ (d) $\cos \theta = \frac{12}{13}$ (e) $\cot \theta = \frac{12}{5}$ 2. (a) i. Length = $\sqrt{5}$ (a) ii. Length = $\sqrt{106}$ (b) i. Gradient = $-\frac{14}{5}$ (b) ii. Gradient = $1$ (c) Coordinates of $M = (6, -1)$