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Taking a wild guess here, but it sounds like you're asked to find |A| and |B| given the magnitude of the resultant A + B, i.e. |A + B| = 50 N, and the sum of the individual magnitudes, |A| + |B| = 70 N.
Recall that the dot product of a vector with itself is equal to the square of that vector's magnitude,
A • A = |A|²
Then
|A + B|² = (A + B) • (A + B)
|A + B|² = (A • A) + 2 (A • B) + (B • B)
|A + B|² = |A|² + 2 (A • B) + |B|²
Since A and B are perpendicular to one another, their dot product is
A • B = 0
So it follows that
|A + B|² = |A|² + |B|²
2500 N² = |A|² + |B|²
Substitute |B| = 70 N - |A| and solve for |A| :
2500 N² = |A|² + (70 N - |A|)²
2500 N² = |A|² + 4900 N² - (140 N) |A| + |A|²
2 |A|² - (140 N) |A| + 2400 N² = 0
|A|² - (70 N) |A| + 1200 N² = 0
|A|² - (70 N) |A| = - 1200 N²
|A|² - (70 N) |A| + 1225 N² = - 1200 N² + 1225 N²
(|A| - 35 N)² = 25 N²
|A| - 35 N = ± 5 N
|A| = 35 N ± 5 N
so that |A| = 30 N or |A| = 40 N. If we fix |A| to be one of these, then |B| will have the other value.
So, the magnitudes are |A| = 30 N and |B| = 40 N.