Question

A plane flies 452 miles north and then 767 miles west. What is the magnitude and direction of the plane's resultant vector?

Answer 1

Answer:

$\displaystyle Approximately\:59°\:at\:a\:magnitude\:of\:approximately\:890\:miles$

Step-by-step explanation:

$\displaystyle \frac{OPPOCITE}{HYPOTENUSE} = sin\:\theta \\ \frac{ADJACENT}{HYPOTENUSE} = cos\:\theta \\ \frac{OPPOCITE}{ADJACENT} = tan\:\theta \\ \frac{HYPOTENUSE}{ADJACENT} = sec\:\theta \\ \frac{HYPOTENUSE}{OPPOCITE} = csc\:\theta \\ \frac{ADJACENT}{OPPOCITE} = cot\:\theta$

We must use trigonometry to help us find the direction of the aeroplane's resultant vector. Do as I do:

$\displaystyle \frac{452}{767} = cot\:x \hookrightarrow cot^{-1}\:\frac{452}{767} = x; 59,488772482...° = x \\ \\ \boxed{59° \approx x}$

Now, we will use the Pythagorean Theorem to find the magnitude of the aeroplane's resultant vector. Do as I do:

$\displaystyle a^2 + b^2 = c^2 \\ \\ 767^2 + 452^2 = c^2 \\ \sqrt{792593} = \sqrt{c^2}; 890,27692321... = c \\ \\ \boxed{890 \approx c}$

Therefore, the direction and magnitude of the aeroplane's resultant vector are approximately eight hundred ninety miles at an angle of elevation of fifty-nine degrees.

I am joyous to assist you at any time.

Answer 2

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:

$\boxed{a^2 = b^2 + c^2}$

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 = $452^2 + 767^2$

⇒ resultant =  $\sqrt{452^2 + 767^2}$

⇒ 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:

$tan (x) =\frac{767}{452}$

⇒ $x = tan^{-1} (\frac{767}{452} )$

⇒ $x = \bf{59.5 \textdegree}$

∴ The plane's resultant vector is 890.3 miles, at an angle of 59.5° west of north .