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Two cyclist X and Y leave town A at the same time. Cyclist X travels at the rate of 10km per hour on a bearing of 049degrees and cyclist Y travels at the rate of 18km on a bearing of 319degrees calculate (a) bearing of cyclist X from Y (b)find the average speed at which cyclist X will get to Y in 4years.

**Comments**

Yes it's in km/h

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please check the question properly... is it 18km or 18km per hour

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a) tanθ = opp/adj

opp = 10

adj = 18

tanθ = 10/18

tanθ = 0.8

θ = tan^{-1}(0.8)

θ = 29.05

Therefore, the bearing of x from y is 90 + 49 - θ

= 180 + 49 - 29.05

= 109.95

b) from pyhtagoras theorem, hyp^{2} = opp^{2} + adj^{2}

hyp = XY

opp = 18

adj = 10

XY^{2} = 18^{2} + 10^{2}

XY = √(18^{2} + 10^{2})

XY = 20.59

speed of XY = distance/time

time = 4 years = 35040 hours

speed of XY = 20.59/35040

= 0.00059km/h **OR **5.1475km/year

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