Two masses `m_(1)` and `m_(2)` are attached with light strings as shown. If the system is in equilibrium, find the value of `theta`

Explanation

Ball of mass `m_(1):`

`T_(1)=m_(1)g`
Joint `A`

By Lami's theorem
`(T_(3))/(sin(90+53^(@)))=(m_(1)g)/(sin(90^(@)))=(T_(3))/(cos53^(@))=m_(1)g`
`rArrT_(3)=m_(1)gcos53^(@)`
`T_(3)=(3m_(1)g)/(5)`
Ball of mass `m_(2):`

`uarr:T_(4)sintheta=m_(2)g+T_(3)sin37^(@)`
`=m_(2)g+(3m_(1)g)/(5)xx(3)/(5)`
`=m_(2)g+(9m_(1)g)/(25)` (i)
`larr:T_(4)costheta=T_(3)cos37^(@)`
`=(3m_(1)g)/(5)xx(4)/(5)=(12m_(1)g)/(25)` (ii)
Dividing (i) by (ii), we get
`tantheta=(m_(2)g+(m_(1)g)/(25))/((12m_(1)g)/(25))=(9m_(1)+25m_(2))/(12m_(1))`