Question

A boy walks 60m on a bearing of 035° and then 300m due east.How far is he from his starting point?

Answer 1

in my diagram, the angle between the two paths is 125°
if x is the distance between them,
x^2 = 60^2 + 300^2 - 2(60)(300)(cos 125°)
= 114,248.7517

x = appr 338 metres

Answer 2

We can solve this problem using the cosine rule and the sine rule.

First, we need to find the angle between the initial bearing and the final eastward direction. To do this, we subtract 90° from the initial bearing of 35°, since east is 90° from north.

Angle = 35° - 90° = -55°

Next, we can use the cosine rule to find the distance between the starting point and the final position:

distance^2 = 60^2 + 300^2 - 2(60)(300)cos(-55°)
distance^2 = 36000 + 90000 + 36000cos(-55°)
distance^2 = 172324.67

distance = sqrt(172324.67)
distance = 415.5m (to 1 decimal place)

Therefore, the boy is 415.5m from his starting point.

Answer 3

That's correct! The distance between the two paths is approximately 338 metres. Good job!