As it is given that 2 kg mass is suspended by 12 cm long thread and then a horizontal force is applied on it so that it remains in equilibrium at 30 degree angle
So here we can use force balance in X and Y directions
now for X direction or horizontal direction we can use
![F = Tsin30](https://img.qammunity.org/2019/formulas/physics/middle-school/tr8bx7yikocn6yagjlxfpizyx5zq13le2d.png)
for vertical direction similarly we can say
![mg = T cos30](https://img.qammunity.org/2019/formulas/physics/middle-school/66pj6h33rwk2cf249zvv2bsfhxjsmzy6kh.png)
so here we first divide the two equations
![(F)/(mg) = (sin 30)/(cos 30)](https://img.qammunity.org/2019/formulas/physics/middle-school/gjmw31y6yo8sr4cf1su3y20v21z53as27r.png)
![(F)/(mg) = tan 30](https://img.qammunity.org/2019/formulas/physics/middle-school/waixca6vewxlm6r4ttswdwqpt1u2kgkpov.png)
![F = mg tan30](https://img.qammunity.org/2019/formulas/physics/middle-school/rr8l023cqmqlfedu0juv66s3ts3d4mqit5.png)
now plug in all values in the above equation
![F = 2 * 9.8 * tan30](https://img.qammunity.org/2019/formulas/physics/middle-school/rxnsyvhcr36577r3i1yw0yfysd1n1w463i.png)
![F = 11.3 N](https://img.qammunity.org/2019/formulas/physics/middle-school/x7ihez9vslzkb6859w8i03qzbqmakj332i.png)
Part b)
now in order to find the tension in the thread we can use any above equation
![F = T sin30](https://img.qammunity.org/2019/formulas/physics/middle-school/sctuza1wpdd3ib8eapuo6ntmsppkmp4hjv.png)
![11.3 = T sin30](https://img.qammunity.org/2019/formulas/physics/middle-school/yabh9yfrwm39gzruqhtu3yuxj6j4i41m0s.png)
![T = (11.3)/(sin30)](https://img.qammunity.org/2019/formulas/physics/middle-school/toljakg99zrkjbh8syc5dblpd0zm0eyvnx.png)
![T = 22.6 N](https://img.qammunity.org/2019/formulas/physics/middle-school/39m01we58d7tqpkaqdy3uboriqv8e66bdh.png)
so tension in the thread will be 22.6 N