A boat seils 10km N 35°E and then 30km N 21°w how far is the boat east and north of the starting point ?

To find out how far the boat is east and north of the starting point, we can break down the distances and directions given.

First, we need to find the eastward and northward components of each leg of the journey:

1. The boat sails 10km N 35°E:
- Eastward component: 10km * cos(35°) ≈ 8.17km
- Northward component: 10km * sin(35°) ≈ 5.74km

2. The boat then sails 30km N 21°W:
- Eastward component: 30km * cos(21°) ≈ 28.02km
- Northward component: 30km * sin(21°) ≈ 10.48km

Now, we can add up the eastward and northward components to find out how far the boat is east and north of the starting point:

Eastward distance = 8.17km + 28.02km ≈ 36.19km
Northward distance = 5.74km + 10.48km ≈ 16.22km

Therefore, the boat is approximately 36.19km east and 16.22km north of the starting point.

Step 1: Convert the directions to unit vectors

- N 35°E:
$$\cos35° \hat{i} + \sin35° \hat{j}$$
- N 21°W:
$$-\cos21° \hat{i} + \sin21° \hat{j}$$

Step 2: Multiply the distances by the unit vectors

- 10 km N 35°E:
$$10(\cos35° \hat{i} + \sin35° \hat{j}) = 8.19\hat{i} + 5.74\hat{j}$$
- 30 km N 21°W:
$$30(-\cos21° \hat{i} + \sin21° \hat{j}) = -28.18\hat{i} + 10.91\hat{j}$$

Step 3: Combine the vectors

$$8.19\hat{i} + 5.74\hat{j} - 28.18\hat{i} + 10.91\hat{j} = -20.00\hat{i} + 16.65\hat{j}$$

Step 4: Find the magnitude and direction of the resultant vector

- Magnitude:
$$\sqrt{(-20.00)^2 + (16.65)^2} km = 26.69 km$$
- Direction:
$$\tan^{-1}\left(\frac{16.65}{-20.00}\right) = -41.24°$$

Therefore, the boat is 20.00 km west and 16.65 km north of the starting point.