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| 1. |
If A,B,C are the angles of a ∆ABC ,show that sin( B+C/2= cosA/2 |
| Answer» Using the angle sum property of a triangle, we get:{tex}sin(\\frac{B+C}{2}){/tex}\xa0=\xa0{tex}sin(\\frac{180^\\circ - A}{2} ){/tex}\xa0[{tex}\\,\\because A + B + C = 180^\\circ{/tex}]or {tex}sin(\\frac{B+C}{2}){/tex}\xa0=\xa0{tex}sin(90^\\circ - \\frac{A}{2}){/tex}But sin and cos are complementary ratios; {tex}sin(90^\\circ-\xa0\\omega) = cos\\,\\omega{/tex}\xa0{tex}\\therefore sin(\\frac{B+C}{2}){/tex}\xa0= {tex}cos(\\frac{A}{2}){/tex}\xa0 | |