Resolving the forces for an object suspended on two strings.

Imagine a following situation: a stationary (not moving) object with a mass of 5 kg is suspended in the air by two strings. The angles between strings and the vertical directions are 15 and 55 degrees. What are tensions T₁ and T₂ in both strings?The first thing to do would be just drawing a simple diagram. I don't think it's possible to add one here, but I drew a little diagram myself! Now, we have to resolve the forces in the horizontal and vertical directions. Because the object is stationary, according to Newton's laws of dynamics we know that adding all the forces will equal 0. Let's start!Vertical forces: we have two vertical components of tension T₁ and T₂ pointing upwards (call it +ve direction) and weight i.e. 5g pointing downwards (let's take g = 9.8). So: T₁cos(15^o) + T₂cos(55^o) - 59.8 = 0  (eqn. 1)Horizontal forces: one component points to the left (-ve direction), the other to the right (+ve direction). Hence: T₁sin(15^o) - T₂sin(55^o) = 0  (eqn. 2)We can see equations 1 and 2 form a set of simultaneous equations. Let's rearrange 2 to get T₁ in terms of T₂: T₁ = T₂sin(55^o) / sin(15^o)  (eqn. 3).Substituting equation 3 into equation 1 gives: T₂sin(55^o)*cot(15^o) + T₂cos(55^o) = 49  (eqn. 4) .From equation 4: T₂ = 13.5 N and hence from equation 3: T₁ = 42.7 N.