(7) 2 sin11°15' is equal to- (A)√2-√2+√2 (B) √2 -

Question

(7) 2 sin11°15' is equal to-(A)√2-√2+√2 (B) √2 -√2-√2 (C) √2+√2+√2/2 (D) √2+√2-√2/2 (8) if 60°+a & 60°-a are roots of sin^2 x+bsinx + c =0 , then -(A)4b^2 + 3=12c(B)4b + 3 =12c(C)4b^2 - 3=-12c(D)4b^2 - 3=12c

Arun · Accepted Answer

Dear Lisa Please post only one question in one thread. sin (60 + a) + sin (60 – a) = – b 2 sin 60 * cos a = – bsqrt (3) * cos a = – b now sin (60 + a) sin (60 – a) = c sin^2 60 – sin^2 a = c sin^2 a = ¾ – c 1 – b^2 / 3 = ¾ – c 1 2 – 4 b^2 = 9 – 12c 4b^2 – 3 = 12C

Trigonometry> (7) 2 sin11°15' is equal to- (A)√2-√2+√2 ...

(7) 2 sin11°15' is equal to-
(A)√2-√2+√2 (B) √2 -√2-√2 (C) √2+√2+√2/2 (D) √2+√2-√2/2

(8) if 60°+a & 60°-a are roots of sin^2 x+bsinx + c =0 , then -
(A)4b^2 + 3=12c
(B)4b + 3 =12c
(C)4b^2 - 3=-12c
(D)4b^2 - 3=12c

Lisha , 7 Years ago

Arun

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Trigonometry> (7) 2 sin11°15' is equal to- (A)√2-√2+√2 ...

(7) 2 sin11°15' is equal to-(A)√2-√2+√2 (B) √2 -√2-√2 (C) √2+√2+√2/2 (D) √2+√2-√2/2 (8) if 60°+a & 60°-a are roots of sin^2 x+bsinx + c =0 , then -(A)4b^2 + 3=12c(B)4b + 3 =12c(C)4b^2 - 3=-12c(D)4b^2 - 3=12c

Lisha , 7 Years ago

Arun

LIVE ONLINE CLASSES

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(7) 2 sin11°15' is equal to-
(A)√2-√2+√2 (B) √2 -√2-√2 (C) √2+√2+√2/2 (D) √2+√2-√2/2

(8) if 60°+a & 60°-a are roots of sin^2 x+bsinx + c =0 , then -
(A)4b^2 + 3=12c
(B)4b + 3 =12c
(C)4b^2 - 3=-12c
(D)4b^2 - 3=12c

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