In any triangle ABC, prove that :

sinA/a = sinB + sinC / b+c = sinB-sinC / b-c​

2 thoughts on “In any triangle ABC, prove that :<br /><br />sinA/a = sinB + sinC / b+c = sinB-sinC / b-c​”

1. In any triangle ABC,

   sinA/a = sinB + sinC / b+c = sinB-sinC / b-c

   Reply

2. In any ∆ABC, using sine rule we say:

   a/sinA = b/sinB = c/sinC, where a, b, c are the sides opposite to angle A, B, and C respectively.

   Let all these be equal to k.

   a/sinA = b/sinB = c/sinC = k

   => a = ksinA ; b = ksinB ; c = ksinC

   Then,

   • a = ksinA

   => 1/k = sinA/a … (1)

   • b + c = ksinB + ksinC

   => b + c = k(sinB + sinC)

   => 1/k = (sinB + sinC)/(b + c) …(2)

   • b – c = ksinB – ksinC

   => b – c = k(sinB – sinC)

   => 1/k = (sinB – sinC)/(b – c) …(3)

   Therefore, from (1), (2) & (3):

   sinA/a=(sinB+sinC)/(b+c)=(sinB-sinC)/(b-c)

   Proved.

   Reply