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?
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).
To find the distance and bearing from the base camp, we can use vector addition and trigonometry.
First, let's convert the bearings to angles in degrees:
032° = 32°
154° = 154°
Now, let's find the x and y components of each leg of the journey using trigonometry.
Leg 1: Distance = 42km, Bearing = 32°
x1 = 42 * cos(32°) ≈ 35.217 km
y1 = 42 * sin(32°) ≈ 21.561 km
Leg 2: Distance = 28km, Bearing = 154°
x2 = 28 * cos(154°) ≈ -7.596 km
y2 = 28 * sin(154°) ≈ 25.751 km
Next, let's add the x and y components to find the final displacement from the base camp.
x_total = x1 + x2 ≈ 35.217 km + (-7.596 km) ≈ 27.621 km
y_total = y1 + y2 ≈ 21.561 km + 25.751 km ≈ 47.312 km
To find the distance from the base camp, we can use the Pythagorean theorem:
distance = sqrt(x_total^2 + y_total^2) ≈ sqrt((27.621 km)^2 + (47.312 km)^2) ≈ sqrt(764.834 km^2 + 2235.065 km^2) ≈ sqrt(3000.899 km^2) ≈ 54.834 km
The man is approximately 54.834 km from his base camp.
Lastly, let's find the bearing from the base camp using trigonometry.
bearing = arctan(y_total / x_total)
bearing ≈ arctan(47.312 km / 27.621 km)
Note: Depending on the calculator or software being used, the bearing angle may be given in the range -180° to +180° or in the range 0° to 360°. For simplicity, let's assume the range is 0° to 360°.
When the arctan is taken, the result is approximately 59.898°.
Since the x and y components are both positive, the bearing is in the first quadrant.
Therefore, the man's bearing from the base camp is approximately 59.898°.