Answer:
The resultant vector is 45.3 N 40.74° North of East
Explanation:
The given vectors can be presented as follows;
35 km at 25 degrees N of E = 35 × cos(25)·[tex]\mathbf{\hat i}[/tex] + 35 × sin(25)·[tex]\mathbf{\hat j}[/tex]
15 km at 10 degrees E of N = 15 × cos(90 - 10)·[tex]\mathbf{\hat i}[/tex] + 15 × sin(90 - 10)·[tex]\mathbf{\hat j}[/tex]
15 km at 10 degrees E of N = 15 × cos(80)·[tex]\mathbf{\hat i}[/tex] + 15 × sin(80)·[tex]\mathbf{\hat j}[/tex]
The sum of the vectors becomes;
∑d = 35 × cos(25)·[tex]\mathbf{\hat i}[/tex] + 15 × cos(80)·[tex]\mathbf{\hat i}[/tex] + 35 × sin(25)·[tex]\mathbf{\hat j}[/tex] + 15 × sin(80)·[tex]\mathbf{\hat j}[/tex]
∑d ≈ 34.33·[tex]\mathbf{\hat i}[/tex] + 29.56·[tex]\mathbf{\hat j}[/tex]
The magnitude of the resultant vector = √(34.33² + 29.56²) ≈ 45.3 N
The direction = tan⁻¹(29.56/34.33) ≈ 40.74° North of East
The resultant vector = 45.3 N 40.74° North of East
