Points P and Q respectively 24m north and 7m east point R. What is the bearing of Q from P to the nearest whole degree?
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Correct Answer: Option E
Explanation:
\(\tan P = \frac{7}{24}\)
\(\tan P = 0.2917\)
\(P = \tan^{-1} (0.2917)\)
= 16.26°
Bearing of Q from P = 180° - 16.26°
= 163.74° \(\approxeq\) 164°
\(\tan P = \frac{7}{24}\)
\(\tan P = 0.2917\)
\(P = \tan^{-1} (0.2917)\)
= 16.26°
Bearing of Q from P = 180° - 16.26°
= 163.74° \(\approxeq\) 164°