A particle of weight 8 N is attached at C to the ends of two light inextensible strings AC and BC - Edexcel - A-Level Maths Mechanics - Question 2 - 2013 - Paper 1

To find the tension in string AC, we start by resolving the forces acting on the particle vertically and horizontally.

1. Resolve vertically: The weight of the particle can be expressed as: $T_A \sin(35^\circ) + T_B \sin(25^\circ) = 8$

2. Resolve horizontally: The tensions in the strings must also balance out: $T_A \cos(35^\circ) = T_B \cos(25^\circ)$

3. By rearranging the horizontal equation, we can express ( T_B ) in terms of ( T_A ): $T_B = T_A \frac{\cos(35^\circ)}{\cos(25^\circ)}$

4. Substitute ( T_B ) back into the vertical equation to solve for ( T_A ):

   $T_A \sin(35^\circ) + T_A \frac{\cos(35^\circ)}{\cos(25^\circ)} \sin(25^\circ) = 8$

5. Solving gives:

   $T_A = 8.4 \text{ N (approximately)}$