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
The correct answer is: E
Explanation
Using pythagoras' theorem:
Hyp\(^2\) = Adj\(^2\) + Opp\(^2\) β 24\(^2\) + 7\(^2\)
Hyp = β625 β 25
\(\tan P = \frac{7}{24}\)
\(\tan P = 0.2917\)
\(P = \tan^{-1} (0.2917)\)
= 16.26Β°
Bearing of Q from P = 180Β° - 16.26Β°
= 163.74Β° \(\approxeq\) 164Β°