If cos a + 2cos b + cos c = 2 then a, b, c are in
- a)2b = a+c
- b)b2 = a x c
- c)a = b = c
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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May 08, 2021 |
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cos A + 2 cos B + cos C = 2
⇒ cos A + cos C = 2(1 – cos B)
⇒ cos A + cos C = 2(1 – cos B)
2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2)
2 sin(B/2) cos((A-C)/2) = 4 sin² (B/2)
cos((A-C)/2) = 2 sin (B/2)
cos((A-C)/2) = 2 cos((A+C)/2)
cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
cos((A-C)/2) = 2 sin (B/2)
cos((A-C)/2) = 2 cos((A+C)/2)
cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)
2 sin(A/2) sin(C/2) = sin(B/2)
⇒ 2 {√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
⇒ 2 {√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac
2(s – b) = b
a + b + c – 2b = b
a + c – b = b
a + c = 2b
a + b + c – 2b = b
a + c – b = b
a + c = 2b
cos A + 2 cos B + cos C = 2⇒ cos A + cos C = 2(1 – cos B)2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2)2 sin(B/2) cos((A-C)/2) = 4 sin² (B/2)cos((A-C)/2) = 2 sin (B/2)cos((A-C)/2) = 2 cos((A+C)/2)cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)2 sin(A/2) sin(C/2) = sin(B/2)⇒ 2 {√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac2(s – b) = ba + b + c – 2b = ba + c – b = ba + c = 2b
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cos A + 2 cos B + cos C = 2⇒ cos A + cos C = 2(1 – cos B)2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2)2 sin(B/2) cos((A-C)/2) = 4 sin² (B/2)cos((A-C)/2) = 2 sin (B/2)cos((A-C)/2) = 2 cos((A+C)/2)cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2)2 sin(A/2) sin(C/2) = sin(B/2)⇒ 2 {√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac2(s – b) = ba + b + c – 2b = ba + c – b = ba + c = 2b
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