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?

  • A 16o
  • B 17o
  • C 73o
  • D 106o
  • E 164o
waec 1988mathematics

The correct answer is E. 164o

Using pythagoras' theorem:

Hyp\(^2\) = Adj\(^2\) + Opp\(^2\) â†’ 24\(^2\) + 7\(^2\)

Hyp = âˆš625 â†’ 25

\(\tan P = \frac{7}{24}\)

\(\tan P = 0.2917\)

\(P = \tan^{-1} (0.2917)\)

= 16.26° 

Bearing of Q from P = 180° - 16.26°

= 163.74° \(\approxeq\) 164°

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