A body walks 60m on a bearing of 035degree and then 300m due east. How far is he from his starting point.
We can use the cosine and sine rules to solve this problem:
First, we can draw a diagram to visualize the situation:
```
A
*
/|
/ |
60/ | 300
/ |
/ |
/ θ |
*------*
B 035°
```
Let's label point A as the final destination of the body and point B as the starting point. We also know that the body walks 60m on a bearing of 035°, so we can add that to our diagram as a line segment:
```
A
*
/|
/ | 60
60/ |
/ |
/ |
/ θ |
*------*
B 035°
```
Next, we know that the body walks 300m due east. We can add that to our diagram as well:
```
A
*
/|
/ | 60
60/ |
/ |
/ | 300
/ θ |
*------*
B 035°
```
Now, we want to find the distance between points A and B, which is the distance the body traveled in total. We can use the cosine rule to find the length of the side connecting A and B:
```
c² = a² + b² - 2ab cos(C)
```
where `a` and `b` are the lengths of the sides adjacent to the angle `C` (the angle between sides `a` and `b`), and `c` is the length of the side opposite to `C`.
In our case, we want to find the length of side `c`, which is the distance between points A and B. We already know the lengths of sides `a` and `b`, which are 60m and 300m, respectively. To find the angle `C`, we can subtract 035° from 180° (since angles on a straight line add up to 180°):
```
C = 180° - 35° = 145°
```
Now we can plug in the values into the cosine rule:
```
c² = a² + b² - 2ab cos(C)
c² = 60² + 300² - 2(60)(300) cos(145°)
c ≈ 319.09
```
Therefore, the distance between points A and B (and the total distance the body traveled) is approximately 319.09m.
To find the distance from the starting point, we can use vector addition.
1. Convert the bearing of 035 degrees into a vector. Since the bearing is measured from the north and east directions, we can break it down into its north and east components using trigonometry.
North component = cos(bearing) * magnitude
= cos(35 degrees) * 60m
≈ 0.819 * 60m
≈ 49.14m (rounded to 2 decimal places)
East component = sin(bearing) * magnitude
= sin(35 degrees) * 60m
≈ 0.574 * 60m
≈ 34.44m (rounded to 2 decimal places)
Therefore, the vector for the first part of the walk is approximately (34.44m, 49.14m).
2. For the second part of the walk, the person walks 300m due east. This can be represented as a vector (300m, 0m).
3. To find the total displacement, we add the two vectors.
Displacement = (34.44m, 49.14m) + (300m, 0m)
= (334.44m, 49.14m)
Therefore, the total displacement from the starting point is approximately (334.44m, 49.14m).
4. Finally, to find the distance from the starting point, we use the Pythagorean theorem.
Distance = sqrt((displacement_x)^2 + (displacement_y)^2)
= sqrt((334.44m)^2 + (49.14m)^2)
≈ sqrt(111789.7936m + 2413.8596m)
≈ sqrt(114203.6532m)
≈ 338.24m (rounded to 2 decimal places)
Therefore, the person is approximately 338.24 meters away from the starting point.