Answer:
The plane's resultant vector is 890.3 miles, at an angle of 59.5° west of north.
Step-by-step explanation:
• To find the magnitude of the resultant vector, we have to use Pythagoras's theorem:
[tex]\boxed{a^2 = b^2 + c^2}[/tex]
where:
a ⇒ hypotenuse (= resultant vector = ? mi)
b, c ⇒ the two other sides of the right-angled triangle (= 452 mil North, 767 mi West).
Using the formula:
resultant² = [tex]452^2 + 767^2[/tex]
⇒ resultant = [tex]\sqrt{452^2 + 767^2}[/tex]
⇒ resultant = 890.3 mi
• To find the direction, we can find the angle (labeled x in diagram) that the resultant makes with the north direction:
[tex]tan (x) =\frac{767}{452}[/tex]
⇒ [tex]x = tan^{-1} (\frac{767}{452} )[/tex]
⇒ [tex]x = \bf{59.5 \textdegree}[/tex]
∴ The plane's resultant vector is 890.3 miles, at an angle of 59.5° west of north .
