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Grade: 12th pass
        
(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
one year ago

Answers : (1)

Arun
24480 Points
							
Dear Lisa
 
Please post only one question in one thread.
 
sin (60 + a) + sin (60 – a) = – b
 
2 sin 60 * cos a = – b
sqrt (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
one year ago
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