To find the direction of the car's resultant vector when a car drives north for 40 miles, then turns and drives at an angle of [tex]\( 30^\circ \)[/tex] west of north for 40 more miles, and with a given resultant vector magnitude of 77.27 miles, we proceed as follows:
1. Identify the vectors and angles:
- The magnitude of each leg of the journey: [tex]\( 40 \)[/tex] miles.
- The angle between the two legs of the journey: [tex]\( 30^\circ \)[/tex].
2. Law of Cosines:
Using the law of cosines to find the angle [tex]\( \theta \)[/tex], we relate the sides of the triangle formed:
[tex]\[
c^2 = a^2 + b^2 - 2ab \cos(\theta)
\][/tex]
where:
- [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are the magnitudes of the two vectors (both are 40 miles).
- [tex]\( c \)[/tex] is the magnitude of the resultant vector (77.27 miles).
3. Calculate the angle [tex]\(\theta\)[/tex]:
Rearranging the law of cosines formula to solve for [tex]\(\cos(\theta)\)[/tex]:
[tex]\[
\cos(\theta) = \frac{a^2 + b^2 - c^2}{2ab}
\][/tex]
Substituting the given values:
[tex]\[
\cos(\theta) = \frac{40^2 + 40^2 - 77.27^2}{2 \cdot 40 \cdot 40}
\][/tex]
[tex]\[
\cos(\theta) = \frac{1600 + 1600 - 5970.8529}{3200}
\][/tex]
[tex]\[
\cos(\theta) = \frac{3200 - 5970.8529}{3200}
\][/tex]
\\
[tex]\[
\cos(\theta) = \frac{-2770.8529}{3200}
\][/tex]
\\
[tex]\[
\cos(\theta) = -0.8665159
\][/tex]
4. Find [tex]\( \theta \)[/tex] in degrees:
Using the inverse cosine function to find [tex]\(\theta\)[/tex]:
[tex]\[
\theta = \arccos(-0.8665159)
\][/tex]
The angle in degrees is:
[tex]\[
\theta = 149.9775050131074^\circ
\][/tex]
5. Interpretation:
The direction of the car's resultant vector is approximately [tex]\( 149.98^\circ \)[/tex] west of north.
Therefore, the direction of the car's resultant vector is approximately [tex]\( 149.98^\circ \)[/tex] west of north.
