Do you have a specific question? Where did this relationship come from? I assume it's part of a larger problem? Do you know how to find components of a vector? Read this: Resolving a Vector Into Components
The statement [itex]T_1= T_2+ T3cos(\theta)[/itex] is only true if, "T_{1}", "T_{2}", and "T_{3}" are the lengths of the vectors shown and the horizontal components sum to 0. (If both T_{1} and T_{2} are horizontal and T_{3} isn't, as shown, the vectors them selves cannot sum to 0.) Since T_{1} and T_{2} are both horizontal, you need to look at the horizontal component of T_{3}. Drop a perpendicular from the tip of T_{3} to the horizontal. You get a right triangle with angle [itex]\theta[/itex] and hypotenuse of length T_{3}. The horizontal component of T_{3}, T_{x then satisifies [itex]cos(\theta)= T_x/T_3[/itex] so [itex]T_x= T_3 cos(\theta)[/itex].}
yes u r right, i was solving a problem on surface tension. For the equilibrium of the drop, the horizontal components must balance each other.
yes, its a part of a larger problem; problem on surface tension. yes i know to find components, BTW i appreciate ur help u always replies to my posts; thanks for ur valuable time
Doc Al's point was "Apply that method (of finding components) and you will answer your own question!"