find a magnitude of the force such that if the act at right angle there resultant is √10N but if the act of 50° the resultant is √13N

Let magnitude of the two forces be x and y.

Resultant at right angle R1 = √15N) and at

60 degrees be R2 = √18N.

Now, R1 = √ (x² + y²) = √15,

R2 = √ (x² + y² + 2xycos50) = √18.

So x² + y² = 15,

and x² + y² + 1.29xy = 18,

therefore 1.29xy = 3,

y = 3/1.29x.

y = 2.33/x

Now, x2 + (2.33/x) 2 = 15,

x² + 5.45/x² = 15

multiply through by x²

x⁴ + 5.45 = 15x²

x⁴ - 15x2 + 5.45 = 0

Now find the roots of the equation, and later y. The two values of x will correspond to the

magnitudes of the two vectors.

Good luck