We can solve this problem using trigonometry and vector addition.
- First, let's draw a diagram:
```
N
|
|
60m @ 035° | x
| /
| /
----*-----+------> E
|
|
|
```
The starting point is marked with a star (*), and we are given that the boy walks 60m on a bearing of 035°. This means he is walking towards the northeast, at an angle of 35° clockwise from north. The distance he walks is not important for now, as we only need to find the final distance from his starting point.
- Next, we can break down the second part of the journey into its horizontal and vertical components. The boy walks 300m due east, so his final position will be to the right (east) of his starting point, with no change in his vertical position:
```
N
|
|
| x -->
| /
| /
----*-----+------> E
|
|
|
```
- To find the distance between the two positions, we can add the two vectors together:
```
N
|
|
| x -->
| / \
| / \
----*-----+----------+------> E
| |
| |
| |
```
We can see that the two vectors form a right-angled triangle. To find the hypotenuse (the final distance from the starting point), we can use the Pythagorean theorem:
```
a^2 + b^2 = c^2
a = 60m
b = 300m
c = ?
c^2 = a^2 + b^2
c^2 = 60^2 + 300^2
c^2 = 36000 + 90000
c^2 = 126000
c = sqrt(126000)
c = 355.9m (rounded to one decimal place)
```
Therefore, the boy is 355.9m from his starting point.
A boy walks 60m on a bearing of 035° and then 300m due east .How far is he from his starting point
4 answers
335
I apologize for the mistake in my previous response. The correct answer is 335m. Here's a corrected explanation:
- The boy walks 60m on a bearing of 035°, which means he moves northeast. To find the horizontal and vertical components of this movement, we can use trigonometry:
```
cos(35°) = adjacent / hypotenuse
adjacent = cos(35°) * 60
adjacent = 49.4m
sin(35°) = opposite / hypotenuse
opposite = sin(35°) * 60
opposite = 34.4m
```
Therefore, the boy moves 49.4m to the east and 34.4m to the north.
- Next, the boy walks 300m due east. This adds to his eastward movement, but does not change his northward movement.
- To find the final distance between the boy's starting point and ending point, we can use the Pythagorean theorem:
```
distance^2 = east^2 + north^2
distance^2 = (60 + 300)^2 + 34.4^2
distance^2 = 36000 + 10816
distance^2 = 46816
distance = sqrt(46816)
distance = 216.4m
```
Therefore, the boy is 216.4m from his starting point. However, we need to add the 49.4m he walked to the east in the beginning. Therefore, the final answer is:
```
216.4m + 49.4m = 265.8m
```
The boy is 265.8m from his starting point. I apologize for any confusion caused by my previous response.
- The boy walks 60m on a bearing of 035°, which means he moves northeast. To find the horizontal and vertical components of this movement, we can use trigonometry:
```
cos(35°) = adjacent / hypotenuse
adjacent = cos(35°) * 60
adjacent = 49.4m
sin(35°) = opposite / hypotenuse
opposite = sin(35°) * 60
opposite = 34.4m
```
Therefore, the boy moves 49.4m to the east and 34.4m to the north.
- Next, the boy walks 300m due east. This adds to his eastward movement, but does not change his northward movement.
- To find the final distance between the boy's starting point and ending point, we can use the Pythagorean theorem:
```
distance^2 = east^2 + north^2
distance^2 = (60 + 300)^2 + 34.4^2
distance^2 = 36000 + 10816
distance^2 = 46816
distance = sqrt(46816)
distance = 216.4m
```
Therefore, the boy is 216.4m from his starting point. However, we need to add the 49.4m he walked to the east in the beginning. Therefore, the final answer is:
```
216.4m + 49.4m = 265.8m
```
The boy is 265.8m from his starting point. I apologize for any confusion caused by my previous response.
338
3 answers