(a) Given that (sin x = frac{5}{13}, 0° < x < 90°), find (frac{cos x – 2sin x}{2tan x}).
(b) A ladder, LA, leans against a vertical pole at a point L which is 9.6metres above the groung. Another ladder, LB, 12 metres long, leans on the opposite side of the pole and at the same point L. If A and B are 10 metres apart and on the same straight line as the foot of the pole, calculate, correct to 2 significant figures, the :
(i) length of ladder LA (ii) angle which LA makes with the ground.
Explanation
(a) (sin x = frac{5}{13})
Using SOHCAHTOA, then Opp = 5 and Hyp = 13
(13^{2} = 5^{2} + Adj^{2})
(169 = 25 + Adj^{2})
(Adj = sqrt{169 – 25} = sqrt{144} = 12)
(cos x = frac{12}{13})
(tan x = frac{5}{12})
(frac{cos x – 2sin x}{2tan x} = frac{frac{12}{13} – 2(frac{5}{13})}{2(frac{5}{12})})
= (frac{frac{2}{13}}{frac{5}{6}})
= (frac{12}{65})
(b)
Taking (Delta LPB),
Using Pythagoras theorem, (/BP/^{2} = /BL/^{2} – /LP/^{2})
(i.e. /BP/ = sqrt{12^{2} – 9.6^{2}})
= (sqrt{144 – 92.16} = sqrt{51.84} = 7.2m)
(/PA/ = 10m – 7.2m = 2.8m)
(/LA/^{2} = x^{2})
(x^{2} = /LP/^{2} + /PA/^{2})
(x^{2} = 9.6^{2} + 2.8^{2})
= (92.16 + 7.84)
(x^{2} = 100 implies x = 10m)
(ii) (tan theta_{y} = frac{9.6}{2.8} = 3.4286)
(theta_{y} = tan^{-1} (3.4286) approxeq 74°)