Answer
The magnitude of the larger pull is 358 N.
The magnitude of the smaller pull is 179 N and it is directed $45.8^{\circ}$ east of north.
Work Step by Step
Let the magnitude of the two pulls be $F$ and $2F$.
We can find the east-west component of the larger pull.
$2F~sin(21.0^{\circ}) = 0.717~F$ toward the west
Since the east-west component of the resultant is zero, the east-west component of the smaller pull is $0.717~F$ to the east. We can assume that the angle $\theta$ of the smaller pull is east of north.
$F~sin(\theta) = 0.717~F$
$\theta = sin^{-1}(0.717) = 45.8^{\circ}$
The sum of the north components of the two pulls is equal to the resultant $460.0~N$.
$2F~cos(21.0^{\circ}) + F~cos(45.8^{\circ})= 460.0~N$
$1.87~F + 0.697~F = 460.0~N$
$F = \frac{460.0~N}{2.567} = 179~N$
$2F = (2)(179~N) = 358~N$
The magnitude of the larger pull is 358 N.
The magnitude of the smaller pull is 179 N and it is directed $45.8^{\circ}$ east of north.
