 Hi and welcome to the session. Let's work out the following question. The question says three forces 5p, 10p and 13p in a plane act on a particle. The angle between any two of them is 120 degree find the magnitude and direction. So let us see the solution to this question. First of all, let R be the resultant of forces 5p, 10p and 13p acting at a point o and let theta be the angle which R makes with the direction 5p. 5p is taken along vector o x and vector o y is perpendicular to vector o x. So we see that components of all forces along vector o x is R cos theta is equal to 5p cos 0 degree plus 10p cos 120 degree plus 13p cos 240 degree. Because these are the angles 5p, 10p and 13p make with respect to vector o x. This is equal to 5p plus 10p into minus 1 by 2 plus 13p into minus 1 by 2 and that is equal to minus 13p by 2. Now components of all forces along vector o y is R sin theta is equal to 5p sin 0 degree plus 10p sin 120 degree plus 13p sin 240 degree. This is equal to 0 plus 10p into root 3 by 2 plus 13p into minus root 3 by 2 that is equal to minus 3 root 3p by 2. Now R square is equal to R square cos square theta plus R square sin square theta this is equal to minus 13p by 2 the whole square plus minus 3 root 3p by 2 the whole square which is further equal to 49p square. Since R square is equal to 49p square this implies that R is equal to square root of 49p square that is equal to 7p. Also tan theta is given by R sin theta over R cos theta that is equal to minus 3 root 3 by 2p divided by minus 13p by 2 and that is equal to 3 root 3 divided by 13. So our answer to this question is the magnitude of the resultant is 7p and the direction that is tan theta is equal to 3 root 3 divided by 13. So this is our answer to this question. I hope that you understood the solution. Have a good time.