 Hello students, let's work out the following problem. It says A, B, C is a given triangle in which forces P, Q and R act along O, A, O, B and O, C where O is the in-center of the triangle and R in equilibrium. Prove that P by cos A by 2 is equal to Q by cos V by 2 is equal to R cos C by 2. So let's now move on to the solution. We are given that O is the in-center of the triangle A, B, C and the forces P, Q and R are acting along O, A, O, B and O, C respectively. The forces P, Q, R are forces acting along O, A, O, B and O, C respectively. Now since O is the in-center of triangle A, B, C, O, A, O, B and O, C are the angle bisectors of angle A, B and C. That means this angle is B by 2, this is B by 2, this is A by 2 and this is also A by 2. This is C by 2 and this is also C by 2 since O, A is the angle bisector of angle A and O, B is the angle bisector of angle B. Similarly O, C is the angle bisector of angle C. Now therefore angle B O, C is equal to 180 degrees minus B by 2 minus C by 2. As we know that sum of three angles of a triangle is 180 degrees. So if we consider triangle B O, C, angle O, B, C is B by 2, angle O, C, B is C by 2. Angle B O, C will be 180 minus B by 2 minus C by 2 and this can be again written as 180 minus B by 2 plus C by 2. Now again this 180 minus 1 by 2 into B plus C 180 minus 1 by 2 into B plus C can be written as 90 minus A. B plus C can be written as 180 minus angle A. Again using the fact that sum of three angles of a triangle is 180 degrees. So angle A, this angle is 180 minus angle B minus angle C. So this is equal to 180 degrees minus 90 degrees plus A by 2 that is 90 degrees plus A by 2. So this angle is 90 degrees plus A by 2. Similarly angle A O, B is equal to 90 plus C by 2 and angle A O, C is equal to 90 degrees plus B by 2. Now since the forces are in equilibrium by Lemmy's theorem, we have P upon sine of angle B O, C is equal to Q upon sine of angle C O, A is equal to R upon sine of angle A O, B. We know that the Lemmy's theorem says that if three forces acting on a particle keep it in equilibrium then each is proportional to the sine of the angle between the other two. So P upon sine of B O, C is equal to Q upon sine of angle C O, A is equal to R upon sine of angle A O, B. Now P upon sine of angle B O, C is 90 plus A by 2 is equal to Q upon sine of angle C O, A that is A O, C which is 90 plus B by 2 is equal to R upon sine of angle A O, B which is 90 plus C by 2. Sine of 90 plus theta is cos theta so it becomes cos A by 2 is equal to Q upon cos B by 2 is equal to R upon cos C by 2 and this is what we had to prove. So this completes the question and the session. Bye for now. Take care. Have a good day.