an archer is practising on a training course and moves a distance of 11m[E]of their original position, which was due south of the target. the archer is now [N28W]. how far is the archer from the target?

Answer 1

To determine the distance of the archer from the target, we need to use the Pythagorean theorem since we have two perpendicular sides of a right triangle (N and W).

First, let's split the displacement of the archer into its components:

Displacement along the N-S direction = 11m[E]
Displacement along the W-E direction = 0m, since the archer moved due South

Now, we can create a right triangle with the displacement as the two sides.

The western component of the displacement is N28W, which implies an angle of 28 degrees with the North direction. Since this angle is measured from the North, it forms a reference angle of 90 - 28 = 62 degrees with the positive x-axis (East).

Using trigonometry, we can determine the displacement along the W-E direction:

cos(62 degrees) = W / (Hypotenuse)
W = cos(62 degrees) * Hypotenuse

Since the displacement along the W-E direction is 0m, the Hypotenuse is equal to W.

Therefore:
W = cos(62 degrees) * W
W = 0.458 * W
0.542W = 0

This implies that W = 0.

Now, we can calculate the displacement along the N-S direction:

sin(62 degrees) = N / (Hypotenuse)
N = sin(62 degrees) * Hypotenuse

N = sin(62 degrees) * 11m[E]
N ≈ 9.9m[N]

Therefore, the archer is about 9.9 meters north of the target.