The velocity of the ship = 22 ∠157° = (22 cos 157°) i+ (22 sin 157°) j
The velocity of current = 5 ∠213° = (5 cos 213°) i+ (5 sin 213°) j
So, the resultant velocity = 22 ∠157° + 5 ∠213°
= (22 cos 157°) i+ (22 sin 157°) j + (5 cos 213°) i+ (5 sin 213°) j
= (22 cos 157° + 5 cos 213°) i + (22 sin 157° + 5 sin 213°) j
= -24.444 i + 5.873 j
= 25.14 ∠166.5°
The correct answer is option (1)
<span>1) 25 knots at 166.5 degree</span>
The coordinates of the drop off would be (-2,2)
Decimal: 1.4 (35 divided by 25 is 1.4)
Percentage: 140% ( move decimal point two spaces to the right)
