Given x = (5)/(13) and 0^ x 90^.

a) Step 1: Find $\cos x$. Given $\sin x = \frac{5}{13}$ and $0^\circ \le x \le 90^\circ$. Using the identity $\sin^2 x + \cos^2 x = 1$: $$ \cos^2 x = 1 - \sin^2 x $$ $$ \cos^2 x = 1 - \left(\frac{5}{13}\right)^2 $$ $$ \cos^2 x = 1 - \frac{25}{169} $$ $$ \cos^2 x = \frac{169 - 25}{169} $$ $$ \cos^2 x = \frac{144}{169} $$ Since $0^\circ \le x \le 90^\circ$, $\cos x$ is positive: $$ \cos x = \sqrt{\frac{144}{169}} = \frac{12}{13} $$ Step 2: Find $\tan x$. Using the identity $\tan x = \frac{\sin x}{\cos x}$: $$ \tan x = \frac{\frac{5}{13}}{\frac{12}{13}} $$ $$ \tan x = \frac{5}{12} $$ Step 3: Substitute the values into the expression $\frac{\cos x - 2 \sin x}{2 \tan x}$. Numerator: $$ \cos x - 2 \sin x = \frac{12}{13} - 2\left(\frac{5}{13}\right) $$ $$ = \frac{12}{13} - \frac{10}{13} $$ $$ = \frac{2}{13} $$ Denominator: $$ 2 \tan x = 2\left(\frac{5}{12}\right) $$ $$ = \frac{10}{12} = \frac{5}{6} $$ Now, divide the numerator by the denominator: $$ \frac{\frac{2}{13}}{\frac{5}{6}} = \frac{2}{13} \times \frac{6}{5} $$ $$ = \frac{12}{65} $$ The value of the expression is $\boxed{\frac{12}{65}}$. b) (i) Calculate $|QP|$. Step 1: Find the angles in $\triangle PQR$. The angle of elevation from