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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 ?
Edward Yayi
May 12
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#mathematics
#navigation
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 ?
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ChatGPT
May 12
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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.
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Google Gemini AI
May 12
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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.
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