To solve for the direction of the deer's resultant vector, follow these steps:
1. Vector Components Breakdown:
- The deer initially runs 88 feet directly to the east.
- Then, the deer turns and runs \( 127 \) feet at an angle of \( 11^{\circ} \) north of west.
2. Coordinate System Setup:
- Consider the east direction as the positive x-axis and the north direction as the positive y-axis.
- Decompose the \( 127 \)-foot run into its northward (y) and westward (x) components.
3. Calculating Westward and Northward Components:
- To find the westward distance (x-component):
[tex]\[
x = 127 \cos(11^{\circ})
\][/tex]
- To find the northward distance (y-component):
[tex]\[
y = 127 \sin(11^{\circ})
\][/tex]
4. Resultant Components:
- The total x-component (westward minus eastward distance):
[tex]\[
\text{Resultant}_x = -88 + x = -88 + 127 \cos(11^{\circ})
\][/tex]
Note that east is positive and west is negative in this coordinate system.
- The northward distance remains the same as the y-component found:
[tex]\[
\text{Resultant}_y = y = 127 \sin(11^{\circ})
\][/tex]
5. Magnitude of the Resultant Vector:
- Using the Pythagorean theorem to calculate the magnitude:
[tex]\[
|\vec{R}| = \sqrt{\text{Resultant}_x^2 + \text{Resultant}_y^2}
\][/tex]
6. Direction of the Resultant Vector:
- The direction angle, \( \theta \), north of west, is determined from the arctangent of the ratio of the northward and westward components:
[tex]\[
\theta = \tan^{-1}\left(\frac{\text{Resultant}_y}{|\text{Resultant}_x|}\right)
\][/tex]
Note that \( \text{Resultant}_x \) is negative (west), so the angle calculated will be in the correct north of west orientation.
Given all the above steps and values:
- The magnitude of the resultant vector \( |\vec{R}| \) is approximately \( 43.95 \) feet.
- The direction \(\theta\) is approximately \( 33.46^{\circ} \) north of west.
Hence, the resultant direction \( \theta \) of the deer's resultant vector is approximately:
[tex]\[ \boxed{33.46^{\circ}} \][/tex]
