1. Correct Answer: b: An angle less than 45° North of East
 2. Explanation:
     ● Step 1: Determine the final position of the woman relative to her starting point.  
           ○ She runs 12 km North, then 6 km South. The net movement in the North-South direction is:
         \[
         12 \, \text{km North} - 6 \, \text{km South} = 6 \, \text{km North}
         \]
           ○ She then runs 8 km East.
     ● Step 2: Calculate the resultant displacement from the starting point.  
           ○ The woman is now 6 km North and 8 km East from her starting point. This forms a right triangle where:
             ○ The North-South leg is 6 km.
             ○ The East leg is 8 km.
     ● Step 3: Determine the direction of the resultant vector.  
           ○ The angle \(\theta\) with respect to the East direction can be calculated using the tangent function:
         \[
         \tan(\theta) = \frac{\text{North-South displacement}}{\text{East displacement}} = \frac{6}{8} = 0.75
         \]
           ○ Calculate \(\theta\) using the arctangent function:
         \[
         \theta = \tan^{-1}(0.75) \approx 36.87^\circ
         \]
     ● Conclusion: The angle \(\theta\) is approximately \(36.87^\circ\), which is less than \(45^\circ\). Therefore, the woman is at an angle less than \(45^\circ\) North of East from her starting point. This corresponds to Option B.