If SinA+Sin2B+Sin3C=3then what is the value of A+2B+3C IS?
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The value of A+2B+3C = 3π/2
Given , SinA+Sin2B+Sin3C=3
The maximum of Sin a (a is the angle ) is equal to 1 in it's domain.
The value of the function Sin a will never exceed 1.
-1 <= Sin a <=1 (for a belongs to R)
In SinA+Sin2B+Sin3C the values of SinA , Sin2B and Sin2C has to be equal to 1 for their sum to be equal to 3.
So, SinA = Sin2B = Sin3C = 1
The sin function is equal to 1 at angle π/2.
=> A = 2B = 3C = π/2
=> A + 2B +3C = 3π/2
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