A boy walks 60m on a bearing of 035° and then 300m due east.How far is he from his starting point?
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
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.
That's correct! The distance between the two paths is approximately 338 metres. Good job!
To find the total distance the boy is from his starting point, we need to use trigonometry.
First, let's break down the given information:
1. The boy walks 60m on a bearing of 035°.
2. After that, he walks 300m due east.
To visualize this, let's draw a diagram:
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W ---------------------- E
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Let's assume the starting point is O. So, the boy walks in the direction of 035° from the north, covering a distance of 60m.
Let's mark this point as A.
```
N
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|O-----A
W ---------------------- E
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S
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Next, the boy walks 300m due east from point A. Let's mark this point as B.
```
N
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|O-----A------B
W ---------------------- E
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S
```
To find the distance between point O and point B, we can use the Pythagorean theorem since we have a right angle triangle AOB.
Using the Pythagorean theorem:
(AB)^2 = (OA)^2 + (OB)^2
We can calculate OA and OB:
OA = 60m (Given distance)
OB = 300m (Given distance)
Substituting OA and OB into the equation:
(AB)^2 = (60)^2 + (300)^2
Calculating:
(AB)^2 = 3600 + 90000
(AB)^2 = 93600
Taking the square root of both sides:
AB = √93600
Calculating:
AB ≈ 305.33
Therefore, the boy is approximately 305.33m away from his starting point.
To find the total distance from the starting point, we can break down the boy's motion into two components: the distance traveled in the northeast direction and the distance traveled due east.
1. Distance traveled on a bearing of 035°:
The boy walks 60m on a bearing of 035°. This means that he is walking in the direction 35° clockwise from due north. We can represent this as a right-angled triangle, where the hypotenuse is the distance traveled and the side adjacent to the angle of 035° represents the eastward movement.
Using trigonometry, we can find the eastward distance (adjacent side) as follows:
Eastward distance = distance traveled * cos(35°)
Eastward distance = 60m * cos(35°)
Eastward distance ≈ 60m * 0.819 (rounded to 3 decimal places)
Eastward distance ≈ 49.14m (rounded to 2 decimal places)
2. Distance traveled due east:
The boy then walks 300m due east, which does not involve any trigonometric calculations.
Total distance from the starting point:
We now add the eastward distance to the distance traveled due east.
Total distance = Eastward distance + Distance due east
Total distance ≈ 49.14m + 300m
Total distance ≈ 349.14m (rounded to 2 decimal places)
Therefore, the boy is approximately 349.14m away from his starting point.