Points P and Q are respectively 24m north and 7m east of 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
Verified Correct Answer: E.
Step-by-Step Mathematical Solution
Using Pythagoras' theorem:
Hyp\(^2\) = Adj\(^2\) + Opp\(^2\) → 24\(^2\) + 7\(^2\)
Hyp = \(\sqrt{625}\) → 25
\(\tan P = \frac{7}{24}\)
tan P = 0.2917
P = tan\(^{-1}\) (0.2917)
= 16.26^\circ
Bearing of Q from P = 180^\circ - 16.26^\circ
= 163.74^\circ \(\approxeq\) 164^\circ
Source: Myschool Past Questions Archive (Record #22046)