(a) Given that \(\sin x = \frac{5}{13}, 0° \leq x \leq 90°\), find \(\frac{\cos x – 2 \sin x }{2\tan x}\).
(b) 
The diagram represents the vertical cross-section of a mountain with height NQ standing on a horizontal ground PRN. If the angles of elevation of the top of the mountain from P and R are 30° and 70° respectively and PR = 500m, calculate, correct to 3 significant figures :
(i) |QP| ; (ii) the height of the mountain.
Explanation
(a) 
\(\sin x = \frac{5}{13}\)
\(AB = \sqrt{13^{2} - 5^{2}} = 12\)
\(\cos x = \frac{12}{13}\)
\(\tan x = \frac{5}{12}\)
\(\frac{\cos x - 2 \sin x}{2 \tan x} = \frac{\frac{12}{15} - 2(\frac{5}{13})}{2(\frac{5}{12})}\)
= \(\frac{\frac{12}{13} - \frac{10}{13}}{\frac{5}{6}}\)
= \(\frac{2}{13} \times \frac{6}{5} = \frac{12}{65}\)
(b) (i) 
< QRP = 180° - 70° = 110° (angles on a straight line)
< RQP = 180° - (110° + 30°) = 40°
\(\frac{PR}{\sin 40°} = \frac{PQ}{\sin 110°}\)
\(\frac{500}{\sin 40°} = \frac{PQ}{\sin 110°}\)
\(PQ = \frac{500 \sin 110°}{\sin 40°} = 730.94m\)
\(\approxeq 731 m\)
(ii) \(\sin 30° = \frac{QN}{PQ} = \frac{QN}{730.94}\)
\(QN = 730.94 \times 0.5 = 365.47 m\)
\(\approxeq 365 m\)