(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 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.
(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.
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Correct Answer: Option n
Explanation:
(a) \(\sin x = \frac{5}{13}\)
Using SOHCAHTOA, then Opp = 5 and Hyp = 13
\(13^{2} = 5^{2} + Adj^{2}\)
\(169 = 25 + Adj^{2}\)
\(Adj = \sqrt{169 - 25} = \sqrt{144} = 12\)
\(\cos x = \frac{12}{13}\)
\(\tan x = \frac{5}{12}\)
\(\frac{\cos x - 2\sin x}{2\tan x} = \frac{\frac{12}{13} - 2(\frac{5}{13})}{2(\frac{5}{12})}\)
= \(\frac{\frac{2}{13}}{\frac{5}{6}}\)
= \(\frac{12}{65}\)
(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) \(\tan \theta_{y} = \frac{9.6}{2.8} = 3.4286\)
\(\theta_{y} = \tan^{-1} (3.4286) \approxeq 74°\)
(a) \(\sin x = \frac{5}{13}\)
Using SOHCAHTOA, then Opp = 5 and Hyp = 13
\(13^{2} = 5^{2} + Adj^{2}\)
\(169 = 25 + Adj^{2}\)
\(Adj = \sqrt{169 - 25} = \sqrt{144} = 12\)
\(\cos x = \frac{12}{13}\)
\(\tan x = \frac{5}{12}\)
\(\frac{\cos x - 2\sin x}{2\tan x} = \frac{\frac{12}{13} - 2(\frac{5}{13})}{2(\frac{5}{12})}\)
= \(\frac{\frac{2}{13}}{\frac{5}{6}}\)
= \(\frac{12}{65}\)
(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) \(\tan \theta_{y} = \frac{9.6}{2.8} = 3.4286\)
\(\theta_{y} = \tan^{-1} (3.4286) \approxeq 74°\)