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If A, B and C are interior angles of triangle ABC then show that cos(B+C/2)=sinA/2 |
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Answer» Hope u ll get it In a triangle, sum of all the interior anglesA + B\xa0+ C = 180°⇒ B\xa0+ C = 180° - A⇒ (B+C)/2 = (180°-A)/2⇒ (B+C)/2 = (90°-A/2)⇒ sin (B+C)/2 = sin (90°-A/2)⇒ sin (B+C)/2 = cos A/2 In a triangle abc we know that all three angles is equal to 180° so angle a +b+c=180. Now shift a on the other side i.e. b+c=180-a next divide both sides by 2 so it gives (b+c) /2= 90-a/2. further on put cos on both sides ---. Cos (b+c)/2= cos 90-a/2 and cos 90-a/2 = sin a/2. So substitute it and we get cos (b+c)/2 = sin a/2 Cos(B+C/2) CAN BE WRITTEN AS Sin(90-B+C/2)...by taking the L. C. M, we will get Sin(180-B+C/2) and we know that in a triangle 180-B+C/2 is equal to A . So, our answer will be SinA/2. |
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