WAEC - Mathematics (1988 - No. 14)
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?
16o
17o
73o
106o
164o
คำอธิบาย
Using pythagoras' theorem:
Hyp\(^2\) = Adj\(^2\) + Opp\(^2\) → 24\(^2\) + 7\(^2\)
Hyp = √625 → 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°