Question

what is the angle between 4N force and 6N force so that their resultant is 8 N ? plzzzz plzzzzz anss​

Answer 1

Given:

Resultant(R) = 8N

Magnitude of first force(P)= 4N

Magnitude of second force(P)=6N

To find:

Angle between forces ($\theta$)

Concept Used:

Parallelogram Law of Addition

Solution:

We know that,

${R }^{2} = {P}^{2} + {Q}^{2} + 2PQcos \theta \\ {8}^{2} = {4}^{2} + {6}^{2} + (4)(6) cos \theta \\ 64 = 16 + 36 + 24cos \theta \\ 64 = 52 + 24cos \theta \\ 24cos \theta = 12 \\ cos \theta = \dfrac{1}{2}$

$\boxed{\red{ \theta = 60°}}$

Answer 2

Answer:

Explanation:

A = 4 N

B= 6 N

c= 10 N

100 = 16 + 36 + 48* cos(theta)

48 = 48*cos(theta)

cos(theta) = 1

theta = 0 degree