Asked by Faith
A man walks due west for 4km. He then changes direction and walks on a bearing of 197° until he is south west his starting point how far is he from his starting point
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Let A be the start (0,0). After walking 4 km due west he is at B = (−4,0). From B he walks a distance d on bearing 197° to C, and C is southwest of A so C = (−t,−t) for some t>0.
Using components (east = d·sinθ, north = d·cosθ) with θ = 197°,
C_x = −4 + d sin197, C_y = d cos197.
Equating to (−t,−t) gives
d(sin197 − cos197) = 4,
so d = 4/(sin197 − cos197) = 4/(cos17° − sin17°) ≈ 6.024 km.
The distance from A to C is AC = √2·t = √2·d·cos17° ≈ 8.15 km.
Answer: about 8.15 km from his starting point.
Using components (east = d·sinθ, north = d·cosθ) with θ = 197°,
C_x = −4 + d sin197, C_y = d cos197.
Equating to (−t,−t) gives
d(sin197 − cos197) = 4,
so d = 4/(sin197 − cos197) = 4/(cos17° − sin17°) ≈ 6.024 km.
The distance from A to C is AC = √2·t = √2·d·cos17° ≈ 8.15 km.
Answer: about 8.15 km from his starting point.
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