Question

A weight of 20 KN supported by two cords, one 3 m long and the other 4m long with points of support 5 m apart find the tensions T1 and T2 in the cords

Answer 1

see diagram.
Perhaps the trianlge ABC a right angle triangle .

As the forces are in equilibrium, we can use triangle law of forces:

T1/sin (β+90) = T2/sin (α+90) = W/sin(α+β)
So    T1/cos β = T2 / cos α = W / sin (α+β)

From the triangle ABC,  sin α / sin β = 4/3
                                   3 cos α + 4 cos β = 5
So    T2/T1 = 4/3

solving the above trigonometric equations, we get β = 36.87 deg    α = 53.13 deg
sinα = 4/5 = 0.8, cos α = 3/5 =0.6 ,  sin β = 3/5 = 0.6    cos β = 4/5 = 0.8

T1/0.6 = T2/0.8 = W/1  = 20 kN

T1 = 12 kN
T2 = 16 kN

Answer 2

T1/sin (β+90) = T2/sin (α+90) = W/sin(α+β)
So    T1/cos β = T2 / cos α = W / sin (α+β)

From the triangle ABC,  sin α / sin β = 4/3
                                   3 cos α + 4 cos β = 5
So    T2/T1 = 4/3
β = 36.87 deg    α = 53.13 deg
sinα = 4/5 = 0.8, cos α = 3/5 =0.6 ,  sin β = 3/5 = 0.6    cos β = 4/5 = 0.
T1/0.6 = T2/0.8 = W/1  = 20 kN
T1 = 12 kN
T2 = 16 kN