In the diagram, /PQ/ = 8m, /QR/ = 13m, the bearing of Q from P is 050° and the bearing of R from Q is 130°.
(a) Calculate, correct to 3 significant figures, (i) /PR/ ; (ii) the bearing of R from P.
(b) Calculate the shortest distance between Q and PR, hence the area of triangle PQR.
(a) Calculate, correct to 3 significant figures, (i) /PR/ ; (ii) the bearing of R from P.
(b) Calculate the shortest distance between Q and PR, hence the area of triangle PQR.
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Correct Answer: Option
Explanation:
(a)(i) < PQR = (180° - 130°) + 50° = 100°
\(PR^{2} = PQ^{2} + QR^{2} - 2(PQ)(QR) \cos < PQR\)
= \(8^{2} + 13^{2} - 2(8)(13) \cos 100\)
= \(233 + 36.118\)
\(PR^{2} = 269.118\)
\(PR = 16.405 m \approxeq 16.4 m\)
(ii) \(\frac{QR}{\sin < QPR} = \frac{PR}{\sin < PQR}\)
\(\frac{13}{\sin < QPR} = \frac{16.4}{\sin 100}\)
\(\sin < QPR = \frac{13 \times \sin 100}{16.4}\)
\(\sin < QPR = 0.7806\)
\(< QPR = 51.32° \approxeq 51.3°\)
The bearing of R from P = 50° + 51.3° = 101.3°
\(\approxeq\) 101°.
(b)
In \(\Delta PQD\),
\(\frac{h}{8} = \sin 51.3\)
\(h = 8 \sin 51.3\)
= \(6.243 m \)
\(\approxeq 6.24 m\)
Area of \(\Delta PQR = \frac{1}{2} \times PR \times h\)
\(\frac{1}{2} \times 6.24 \times 16.4 = 51.166 m^{2}\)
\(\approxeq 51.2 m^{2}\)
(a)(i) < PQR = (180° - 130°) + 50° = 100°
\(PR^{2} = PQ^{2} + QR^{2} - 2(PQ)(QR) \cos < PQR\)
= \(8^{2} + 13^{2} - 2(8)(13) \cos 100\)
= \(233 + 36.118\)
\(PR^{2} = 269.118\)
\(PR = 16.405 m \approxeq 16.4 m\)
(ii) \(\frac{QR}{\sin < QPR} = \frac{PR}{\sin < PQR}\)
\(\frac{13}{\sin < QPR} = \frac{16.4}{\sin 100}\)
\(\sin < QPR = \frac{13 \times \sin 100}{16.4}\)
\(\sin < QPR = 0.7806\)
\(< QPR = 51.32° \approxeq 51.3°\)
The bearing of R from P = 50° + 51.3° = 101.3°
\(\approxeq\) 101°.
(b)
In \(\Delta PQD\),
\(\frac{h}{8} = \sin 51.3\)
\(h = 8 \sin 51.3\)
= \(6.243 m \)
\(\approxeq 6.24 m\)
Area of \(\Delta PQR = \frac{1}{2} \times PR \times h\)
\(\frac{1}{2} \times 6.24 \times 16.4 = 51.166 m^{2}\)
\(\approxeq 51.2 m^{2}\)