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

Correct answer is E

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°