(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.

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(a) Given that $\sin x = \frac{5}{13}, 0 ° < x < 90 °$ , find \(\frac{\cos ...

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(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.

Explanation

(a)

\sin x = \frac{5}{13}

Using SOHCAHTOA, then Opp = 5 and Hyp = 13

13^{2} = 5^{2} + A d j^{2}

169 = 25 + A d j^{2}

A d j = \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

Δ L P B

,

Using Pythagoras theorem,

/ B P /^{2} = / B L /^{2} - / L P /^{2}

i . e . / B P / = \sqrt{12^{2} - {9.6}^{2}}

\sqrt{144 - 92.16} = \sqrt{51.84} = 7.2 m

/ P A / = 10 m - 7.2 m = 2.8 m

/ L A /^{2} = x^{2}

x^{2} = / L P /^{2} + / P A /^{2}

x^{2} = {9.6}^{2} + {2.8}^{2}

92.16 + 7.84

x^{2} = 100 ⟹ x = 10 m

(ii)

\tan θ_{y} = \frac{9.6}{2.8} = 3.4286

θ_{y} = \tan^{- 1} (3.4286) ≊ 74 °

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