Question

A helicopter takes off from its home base in travels 33 miles East and then 20 miles North. The result of the trip can be described by the vector vector <33, 20>. What are the magnitude and the direction of the flight vector? Round to the nearest 10th.

Answer 1

Answer:

38.6 miles

31.2° with respect to East in anticlockwise direction.

Step-by-step explanation:

A helicopter takes off from its home base A and travels 33 miles East to B and then 20 miles North to C.

Now, if we consider AB and BC as two vectors then the resultant vector will be the AC vector which is the hypotenuse of the right triangle ABC.

Now, Magnitude of AC vector is $\sqrt{33^{2} + 20^{2}} = 38.6$ miles (Rounded to the nearest tenth)

Now, the direction of this vector will be given by the angle that AC makes with the East direction.

Now, $\tan x = \frac{BC}{AB} = \frac{20}{33} = 0.606$

⇒ $x = \tan^{-1}(0.606) = 31.2$ degrees

Therefore, the direction of vector AC is 31.2° with respect to East in anticlockwise direction. (Answer)