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 $P(3, 2)$ and $Q(4, 1)$ 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 Pythagoras theorem to find $\cos \theta$: $$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$$ 3. Calculate $\tan \theta$: $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$$ 4. Calculate $\csc \theta$ (reciprocal of $\sin \theta$): $$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{5}{13}} = \frac{13}{5}$$ 5. $\sin \theta$ is given as $\frac{5}{13}$. 6. $\cos \theta$ found as $\frac{12}{13}$. 7. Calculate $\cot \theta$ (reciprocal of $\tan \theta$): $$\cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{5}{12}} = \frac{12}{5}$$ --- ### Step 2: Lengths of lines Length formula between points $(x_1, y_1)$ and $(x_2, y_2)$: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ (a)(i) Between $(3, -5)$ and $(4, -3)$: $$d = \sqrt{(4 - 3)^2 + (-3 + 5)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$$ (a)(ii) Between $(-2, 7)$ and $(3, -2)$: $$d = \sqrt{(3 + 2)^2 + (-2 - 7)^2} = \sqrt{5^2 + (-9)^2} = \sqrt{25 + 81} = \sqrt{106}$$ --- ### 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 dividing line in ratio Point $M$ divides $PQ$ in ratio $3 : -2$. Coordinates formula for ratio $m:n$: $$M = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$$ Here, $P(3, 2)$, $Q(4, 1)$, $m=3$, $n=-2$: $$x_M = \frac{3 \times 4 + (-2) \times 3}{3 + (-2)} = \frac{12 - 6}{1} = 6$$ $$y_M = \frac{3 \times 1 + (-2) \times 2}{3 + (-2)} = \frac{3 - 4}{1} = -1$$ So, $M = (6, -1)$. --- ### Final answers: (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 = $\sqrt{5}$ 2.(a)(ii) Length = $\sqrt{106}$ 2.(b)(i) Gradient = $-\frac{14}{5}$ 2.(b)(ii) Gradient = $1$ 2.(c) Coordinates of $M = (6, -1)$