(a) Given that \(\sin x = \frac{5}{13}, 0° < x < 90°\), find \(\frac{\cos x - 2\sin x}{2\tan x}\). (b) A ladder, LA, leans against a vertical pole at a poin...

Answer Details

(a) Given \(\sin x = \dfrac{5}{13}\) with \(0^\circ < x < 90^\circ\). Using a 5-12-13 right triangle (\(\sqrt{13^2 - 5^2} = 12\)):

\[\cos x = \frac{12}{13}, \qquad \tan x = \frac{5}{12}.\]

Now evaluate \(\dfrac{\cos x - 2\sin x}{2\tan x}\):

\[\cos x - 2\sin x = \frac{12}{13} - \frac{10}{13} = \frac{2}{13}, \qquad 2\tan x = \frac{10}{12} = \frac{5}{6}.\]

\[\frac{\cos x - 2\sin x}{2\tan x} = \frac{2/13}{5/6} = \frac{2}{13}\times\frac{6}{5} = \frac{12}{65}.\]

(b) The point \(L\) is 9.6 m up the vertical pole, with foot \(F\). Ladder \(LB = 12\) m reaches the ground at \(B\); \(A\) and \(B\) are 10 m apart on opposite sides of the pole.

Horizontal distance \(FB\):

\[FB = \sqrt{12^2 - 9.6^2} = \sqrt{144 - 92.16} = \sqrt{51.84} = 7.2\text{ m}.\]

Since \(A\) and \(B\) are on opposite sides, \(FA + FB = 10\), so \(FA = 10 - 7.2 = 2.8\text{ m}\).

(i) \(LA = \sqrt{FA^2 + 9.6^2} = \sqrt{2.8^2 + 9.6^2} = \sqrt{7.84 + 92.16} = \sqrt{100} = 10\text{ m}\) (2 s.f.).

(ii) Let \(\theta\) be the angle \(LA\) makes with the ground:

\[\tan\theta = \frac{9.6}{2.8} = 3.4286 \Rightarrow \theta = 73.7^\circ \approx 74^\circ.\]