Question

Two equal forces (P each) act at a point inclined to each other at an angle of 120. The magnitude of their resultant is ans.P

Answer 1

$resultant = \sqrt{ {x}^{2} + {y}^{2} + 2xy \cos(a) }$
where x and y are Magnitudes of two vectors and a is the angle between them.So resultant =
$\sqrt{ {p}^{2} + {p}^{2} + 2(p)(p) \cos(120) }$
$= \sqrt{2 {p}^{2} + 2 {p}^{2} ( - \frac{1}{2}) } = \sqrt{ {p}^{2} } = p.$

Answer 2

we know, force is a vector quantity. so, it follows parallelogram law for addition of two vectors.
Let two forces $F_1$ and $F_2$ are given and angle between them is given $\theta$
then, $F_{net}=\sqrt{F_1^2+F_2^2+2F_1.F_2cos\theta}$


now come to the question,
two forces (each of magnitude P) act at a point inclined each other at an angle of 120°.
then, magnitude of their resultant,
$P_{net}=\sqrt{P^2+P^2+2P.Pcos120^{\circ}}$
$P_{net}=\sqrt{P^2+P^2+2P^2\frac{-1}{2}}$
$P_{net}=\sqrt{P^2+P^2-P^2}=P$

hence, magnitude of their resultant is P