two mountain bikers leave from the same parking lot and head in opposite directions on two different trails. the first rider goes 8km due east, then rides due south for 15km. The second rider goes 8km due west, then changes directions and rides 20 degrees west of due north for 15km. Both riders have been traveling for 23km, but which one is further from the parking lot?
Answers
Applying the required rule or theorem, it can be concluded that the second biker is farther from the parking lot. The distance of the bikers to the parking lot are:
i. First biker = 17.0 km
ii. Second biker = 20.22 km
The path of travel of both bikers would form a triangle. Applying the Pythagoras theorem to the path of the first biker would give his distance from the starting point. While applying the cosine rule to the path of second rider would gives his distance to the starting point.
Thus,
a. To determine the distance of the first biker from the parking lot.
Let the required distance be represented by x. Applying the Pythagoras theorem, we have:
[tex]hyp^{2}[/tex] = [tex]adj 1^{2}[/tex] + [tex]adj 2^{2}[/tex]
[tex]x^{2}[/tex] = [tex]8^{2}[/tex] + [tex]15^{2}[/tex]
= 64 + 225
= 289
x = [tex]\sqrt{289}[/tex]
= 17
x = 17 km
Thus, the first biker is 17.0 km from the starting point.
b. To determine the distance of the second biker from the parking lot.
Let the required distance be represented by x. So that applying the cosine rule, we have:
[tex]c^{2}[/tex] = [tex]a^{2}[/tex] + [tex]b^{2}[/tex] - 2ab Cos θ
[tex]x^{2}[/tex] = [tex]8^{2}[/tex] + [tex]15^{2}[/tex] - 2(15*8) Cos (180 - 20)
= 64 + 225 - 240 Cos 160
= 289 - 240 * -0.5
[tex]x^{2}[/tex] = 289 + 120
= 409
x = [tex]\sqrt{409}[/tex]
= 20.2237
x = 20.22 km
Thus, the second biker is 20.22 km from the starting point.
Therefore, the second biker is farther from the parking lot.
A sketch of the path of travel for the two bikers is attached for more clarifications.
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{
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