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Two cyclists leave from the same starting point. One cyclist travels due west while the other travels on a bearing of 202°. After traveling for 18 km, the second cyclist is due south of the first cyclist. How far (to the nearest meter) has the first cyclist traveled?

a. 18 km
b. 18√3 km
c. 18√2 km
d. 36 km

1 Answer

2 votes

Final answer:

Two cyclists start at the same point, one heading due west and the other on a bearing of 202°. By considering the bearing angle and using trigonometric functions, we calculate the distance the first cyclist traveled to be approximately 16.725 km, but this does not match any of the provided answer choices. Option A is the nearest correct answer.

Step-by-step explanation:

The question involves the analysis of a geometry problem related to displacement, and it can be solved using fundamental trigonometric concepts. The scenario describes two cyclists who start at the same point, with one cyclist heading due west and the other traveling on a bearing of 202°. After the second cyclist travels for 18 km and finds themselves due south of the first cyclist, we are tasked with determining the distance the first cyclist has traveled.

To solve this, we can use a right-angled triangle, with the first cyclist's westward trajectory forming one leg and the second cyclist's path forming the hypotenuse. The bearing of 202° suggests that the second cyclist is traveling 22° south of due west, forming the angle at the starting point. Using trigonometric functions, particularly the cosine for adjacent side calculation, we can determine the westward distance of the first cyclist.

Let the distance traveled by the first cyclist be x.

Cos(22°) = x / 18

x = Cos(22°) × 18

x ≈ 16.725 km (rounded to the nearest meter)

Therefore, the first cyclist has traveled approximately 16,725 meters or 16.725 km due west.

However, none of the provided options (a. 18 km, b. 18√3 km, c. 18√2 km, d. 36 km) match our calculated distance. Thus, there must be an error in the options listed or in the question's premise itself.

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