A man prospecting for oil in the desert leaves his camp and arrives 42km on a bearing of 032°. He then drives 28km on a bearing of 154°. How far is he then from his base camp and what is his bearing from it?

Answer 1

Using trigonometry and vector addition:

1. Convert the two bearings to Cartesian coordinates:

42 km on a bearing of 032° = (42 cos 32°, 42 sin 32°) ≈ (35.75, 21.95)

28 km on a bearing of 154° = (28 cos 154°, 28 sin 154°) ≈ (-16.77, 25.23)

Note: The negative x-component for the second vector indicates it points to the left (west) of the first vector.

2. Add the two vectors to find the resultant displacement:

(35.75, 21.95) + (-16.77, 25.23) ≈ (18.98, 47.18)

This means the man is approximately 18.98 km east and 47.18 km north from his base camp.

3. Find the magnitude and direction of the resultant vector:

magnitude = sqrt((18.98)^2 + (47.18)^2) ≈ 50.72 km

direction = atan(47.18/18.98) ≈ 68.7°

This means the man is approximately 50.72 km away from his base camp, and his bearing from it is approximately 068° (east of north).