5. If A, B and C are interior angles of a triangle ABC, then show that.Sin (B+C÷2)=COS A÷2
Answers
Answer:A+B+C=180
A+B+C=180B+C=180-A
A+B+C=180B+C=180-AB+C/2=90-A/2
A+B+C=180B+C=180-AB+C/2=90-A/2Sin(B+C/2)=Sin(90-A/2)
A+B+C=180B+C=180-AB+C/2=90-A/2Sin(B+C/2)=Sin(90-A/2)Sin(B+C/2)=CosA/2
A + B + C = 180°
A + B + C = 180° Therefore ,
A + B + C = 180° Therefore ,B + C = 180 - A
A + B + C = 180° Therefore ,B + C = 180 - A (B + C) / 2 = (180 - A ) / 2
A + B + C = 180° Therefore ,B + C = 180 - A (B + C) / 2 = (180 - A ) / 2(B + C) / 2 = 90 - (A / 2)
A + B + C = 180° Therefore ,B + C = 180 - A (B + C) / 2 = (180 - A ) / 2(B + C) / 2 = 90 - (A / 2)Hence R.H.S. :- cos (A / 2) = sin [ 90 - (A / 2) ]
A + B + C = 180° Therefore ,B + C = 180 - A (B + C) / 2 = (180 - A ) / 2(B + C) / 2 = 90 - (A / 2)Hence R.H.S. :- cos (A / 2) = sin [ 90 - (A / 2) ]= sin [ (B+C) / 2 ] ----- As 90 - (A / 2) = 2 (B+C) / 2 ]
A + B + C = 180° Therefore ,B + C = 180 - A (B + C) / 2 = (180 - A ) / 2(B + C) / 2 = 90 - (A / 2)Hence R.H.S. :- cos (A / 2) = sin [ 90 - (A / 2) ]= sin [ (B+C) / 2 ] ----- As 90 - (A / 2) = 2 (B+C) / 2 ] Therefore ,
A + B + C = 180° Therefore ,B + C = 180 - A (B + C) / 2 = (180 - A ) / 2(B + C) / 2 = 90 - (A / 2)Hence R.H.S. :- cos (A / 2) = sin [ 90 - (A / 2) ]= sin [ (B+C) / 2 ] ----- As 90 - (A / 2) = 2 (B+C) / 2 ] Therefore , cos (A / 2) = sin [ (B+C) / 2 ]
A + B + C = 180° Therefore ,B + C = 180 - A (B + C) / 2 = (180 - A ) / 2(B + C) / 2 = 90 - (A / 2)Hence R.H.S. :- cos (A / 2) = sin [ 90 - (A / 2) ]= sin [ (B+C) / 2 ] ----- As 90 - (A / 2) = 2 (B+C) / 2 ] Therefore , cos (A / 2) = sin [ (B+C) / 2 ]Hence proved.