A person walks 25.0 degrees north of east for 3.80 km. How far would the person walk due north and due east to arrive at the same location? I did the Sin of 25 but the answer was wrong
You didn't solve for the north and east directions. It isn't sin 25, it's sin 25 = north/3.8 or 3.8*sin 25 = 0.423 x 3.8 = 1.61 km north.
cos 25 = east/3.8
3.8*cos 25 = 0.906*3.8 = 3.44 km east.
To find out how far a person would walk due north and due east to arrive at the same location, we can use trigonometry.
Let's assume the person starts at point A and ends up at point B after walking 3.80 km at an angle of 25.0 degrees north of east. We need to find the distances AB' and AB", where B' is the point directly north of B and B" is the point directly east of B.
First, let's find the distance AB' by considering the northward direction. We can use the sine function since we have the hypotenuse value AB and the angle of 25.0 degrees.
sin(25.0°) = AB' / AB
Rearranging the equation to solve for AB':
AB' = AB * sin(25.0°)
Substituting the given values:
AB' = 3.80 km * sin(25.0°)
Calculating AB':
AB' ≈ 1.60 km
Next, let's find the distance AB" by considering the eastward direction. We can use the cosine function since we have the hypotenuse value AB and the angle of 25.0 degrees.
cos(25.0°) = AB" / AB
Rearranging the equation to solve for AB":
AB" = AB * cos(25.0°)
Substituting the given values:
AB" = 3.80 km * cos(25.0°)
Calculating AB":
AB" ≈ 3.42 km
Therefore, the person would need to walk approximately 1.60 km due north (AB') and approximately 3.42 km due east (AB") to arrive at the same location.
Note: Ensure that your calculator is in degree mode when calculating trigonometric functions.