SOLUTION: A man wandering in the desert walks 2.3 miles in the direction S 31° W. He then turns 90° and walks 3.6 miles in the direction N 59° W. At that time, how far is he from his star
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Question 1145817: A man wandering in the desert walks 2.3 miles in the direction S 31° W. He then turns 90° and walks 3.6 miles in the direction N 59° W. At that time, how far is he from his starting point, and what is his bearing from his starting point?
(Round your answers to the nearest whole number.)
distance _________mi
bearing S ________° W Answer by Theo(13342) (Show Source):
he walks 2.3 miles in the direction of south 31 degrees west.
he then walks 3.6 miles in the direction of north 59 degrees west.
that forms a right angle at point B.
he starts at point A and finishes at point C.
that forms a right triangle ABC.
one leg of the right triangle is 2.3 miles and the other leg of the right triangle is 3.6 miles.
the hypotenuse of the right triangle is srt(3.6^2 + 2.3^2) = sqrt(18.25) miles.
that's the distance from the starting point (A) to the end point (C) which is sqrt(18.25) miles long = 4.272001873 miles = 4 miles rounded to the nearest whole number.
angle x is the same as angle CAB of the right triangle.
sin(x) = opp/hyp = 3.6 / sqrt(18.25).
angle x = arcsin (3.6 / sqrt(18.25)) = 57.42594287 degrees = 57 degrees rounded to the nearest whole number.
angle x + 31 = 57 + 31 = 88 degrees.
the bearing from the starting point (A) to the end point (C) is therefore south 88 degrees west.
