(a) Given that (sin x = frac{5}{13}, 0° leq x leq 90°), find (frac{cos x – 2 sin x }{2tan 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)