in triangle ABC ,write cos(B+C)/2 in terms of angle A.
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Answered by
517
A+B+C = 180 (ASP of trianlge)
~ B + C = 180 - A
~ cos (B + C) / 2 = cos (180 - A) / 2
~ cos (180/2 - A/2)
~ cos (90 - A/2)
~ sin (A/2)
Therefore, cos (B + C/2) = sin (A/2)
~ B + C = 180 - A
~ cos (B + C) / 2 = cos (180 - A) / 2
~ cos (180/2 - A/2)
~ cos (90 - A/2)
~ sin (A/2)
Therefore, cos (B + C/2) = sin (A/2)
Answered by
88
Answer:
A+B+C = 180 (ASP of trianlge)
~ B + C = 180 - A
~ cos (B + C) / 2 = cos (180 - A) / 2
~ cos (180/2 - A/2)
~ cos (90 - A/2)
~ sin (A/2)
Therefore, cos (B + C/2) = sin (A/2)tep-by-step explanation:
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