(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-sec...
Question 1 Report
(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.
Answer Details
(a) Given \(\sin x=\dfrac{5}{13}\) with \(0^\circ\le x\le 90^\circ\), evaluate \(\dfrac{\cos x-2\sin x}{2\tan x}\).
Using a \(5\)-\(12\)-\(13\) right triangle (adjacent \(=\sqrt{13^2-5^2}=\sqrt{144}=12\)):
\[\cos x=\frac{12}{13},\qquad \tan x=\frac{5}{12}.\]Substitute:
\[\frac{\cos x-2\sin x}{2\tan x}=\frac{\dfrac{12}{13}-2\cdot\dfrac{5}{13}}{2\cdot\dfrac{5}{12}}=\frac{\dfrac{12-10}{13}}{\dfrac{10}{12}}=\frac{\dfrac{2}{13}}{\dfrac{5}{6}}.\]\[=\frac{2}{13}\times\frac{6}{5}=\mathbf{\frac{12}{65}}.\](b) \(NQ\) is the vertical height of the mountain on horizontal ground \(PRN\) (with \(R\) between \(P\) and \(N\)). The angle of elevation of the top \(Q\) is \(30^\circ\) from \(P\) and \(70^\circ\) from \(R\), and \(PR=500\text{ m}\).
(i) \(|QP|\). In triangle \(PQR\), \(\angle QPR=30^\circ\). Since \(\angle QRN=70^\circ\) is exterior to the triangle at \(R\), the interior angle is
\[\angle QRP=180^\circ-70^\circ=110^\circ,\]\[\angle PQR=180^\circ-30^\circ-110^\circ=40^\circ.\]By the sine rule:
\[\frac{|QP|}{\sin\angle QRP}=\frac{|PR|}{\sin\angle PQR}\]\[|QP|=\frac{500\times\sin 110^\circ}{\sin 40^\circ}=\frac{500\times0.9397}{0.6428}\approx\mathbf{731\text{ m}}.\](ii) Height of the mountain \(NQ\). In right-angled triangle \(QNP\) (right angle at \(N\)), with \(\angle QPN=30^\circ\):
\[NQ=|QP|\sin 30^\circ=730.95\times0.5\approx\mathbf{365\text{ m}}.\](Check via triangle \(QRN\): \(QR=\dfrac{500\sin30^\circ}{\sin40^\circ}=388.9\text{ m}\), and \(NQ=QR\sin70^\circ=388.9\times0.9397\approx365\text{ m}\).)
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