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suppose the sum of 2 angles A and B =C where C is between -90 to 90 so A + B = C sin(A+B) = sinC sin(A+B) =sin (180-C) A + B = 180 – C A + B + C = 180 plz tell me my mistake as we all know that this thing is not at all true

suppose 
the sum of 2 angles A and B =C
where C is between -90 to 90
so
      A + B = C
\Rightarrow sin(A+B) = sinC
\Rightarrow sin(A+B) =sin (180-C)
\Rightarrow A + B = 180 – C
\Rightarrow A + B + C = 180
plz tell me my mistake as we all know that this thing is not at all true
 
 

Grade:11

1 Answers

Ajay
209 Points
7 years ago
Posting again as some errors in previois answer
 
 sin x = sin y implies x = nπ + (– 1)n y, where n ∈ Z.
 
in Your case Sin (A+B) = Sin(180-C) implies
A+B = 180-C for n = 0
This solution is not correct as explained
Since C -90 implies 180-C  > 90.
But A+B  = C is less than 90 hence this solution is discarded
 
 
The correct solution to this equation is
A+B = 180 – (180-C)  = C for n  =1
 

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