Step 1: Identify the quadrant using signs.
Departure \(+70\) m \(\Rightarrow\) \(\text{East}\).
Latitude \(-130\) m \(\Rightarrow\) \(\text{South}\).
Hence the line lies in the \(\text{SE}\) quadrant (reduced bearing \(= S\,\theta\,E\)).
Step 2: Compute the reduced bearing angle.
\[
\tan\theta = \frac{|\text{dep}|}{|\text{lat}|}=\frac{70}{130}=0.53846
\Rightarrow
\theta \approx \arctan(0.53846)\approx 28.3^\circ.
\]
Step 3: Convert reduced bearing to whole circle bearing.
For the SE quadrant, \(\text{WCB} = 180^\circ - \theta\).
Thus,
\[
\text{WCB} \approx 180^\circ - 28.3^\circ = 151.7^\circ \approx \boxed{152^\circ}.
\]
Ravi had _________ younger brother who taught at _________ university. He was widely regarded as _________ honorable man.
Select the option with the correct sequence of articles to fill in the blanks.