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

look at the diagram

resolve t1 and t2 into horizontal and vertical components
t1 cos a+ t2 cos b=mg
and t1 sin a=t2 sin b
sin a=4/5 sin b= 3/5
4t1=3t2
take t1=3x and t2=4x
t1 cos a + t2 cos b=mg=20KN
3x (3/5)+4x(4/5)=20KN
5x=20
x=4
therefore t1= 12 KN t2 = 16 KN
sorry next time i will do the diagram in a nice way

 

Answer 2

let 3m cord make an angle t with horizontal and have tension T1.  3m, 4m, 5m long cords make a right angle triangle with 90 degrees at junction of T1 and T2. So T2 makes 180 - t - 90 = 90-t with horizontal.
         T1 cos t  = T2 cos 90-t   => T1 = T2 sin t  / cos t          ---  equation 1

In the triangle formed by two points of support and junction of T1 & T2, we have
          3 sin t + 4 cos t = 5  (side joining two supports, its a right angle triangle)
           divide by cos t => 3 tan t + 4 = 5 sec t
           9 tan t * tan t + 16 + 24 tan t = 25 sec t * sec t  = 24 ( 1 + tan t * tan t)
         tan t  = 3/4  by solving quadratic equation, 

          T1 = T2 * 3 /4.     
         T1 * T1 + T2 * T2 + 2 T1 T2 cos 90 = 20 * 20
         T2 * T2 (3 * 3 / 4 * 4) + T2 * T2 = 400  => T2 * T2 = 400 * 16 / 25 = 256
  
       T2 = 16 KN  and T1 = 12 KN