La-2l + lb-3l + lc-4l = 0, find the value of a + B + C

La-2l + lb-3l + lc-4l = 0, find the value of a + B + C

la-2l+lb-3l+lc-4l=0
a=2 b=3 c=4
a+b+c
=9

Given la-2l + LB + 3L + lc-4l = 0, find the value of a + 2lbl + 3C?

Given the internal angle La LB LC 2lc = La + lb, la-lb = 40 ° of triangle, find the degree of La LB LC
L is the angle

2lc = La + LB Formula 1
La-lb = 40 equation 2
La + LB + LC = 180 formula 3
From Formula 1 and formula 3, 3lc = 180, so LC = 60
So La + LB = 120 equation 4
Equation 4 plus equation 2 gives 2La = 160, so La = 80
LB=LA-40=40
To sum up, La = 80, LB = 40, LC = 60