if abc are the interior angels of a triangle ABC, show that :sin C+B/2=cosA/2
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Answer:
Step-by-step explanation:
Given ,
A,B,C are angles of triangle ABC
So ,
A + B + C = 180° —(1) ( angle sum property)
Divide equation (1) by 2
A/2 + B/2 + C/2 = 90°
=> A/2 + B+C/2 = 90°
=> B+C/2 = 90°- A/2 —(2)
Apply sin ratio on both sides of the equation (2)
Sin (B+C/2) = sin (90-A/2)
Sin (B+C/2) = Cos A/2 [sin(90-A) = cos A]
Hence proved
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