In △ ABC, if A is an acute angle, b=6 ,c=9 and sin ⁡A= 2 149, then 3a ( cos ⁡B+ cos ⁡C )=

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Correct answer: B. (b) b=6 ,c=9 sin ⁡A= 2 149 ∴ cos ⁡A= 1- 5681= 59 Since cos ⁡A= b^2+c^2-a^22 bc ⇒ 59= 36+81-a^22 (6 ) (9 ) ∴a^2=573a ( cos ⁡B+ cos ⁡C )=3a( a^2+c^2-b^22 ac+ a^2+b^2-c^22 ab) =3a ( ba^2+bc^2-b^3+a^2c+b^2c-c^32 abc ) = 32 bc(a^2 (b+c )+ bc (b+c )-(b^3+c^3)) = 32 bc (b+c )(a^2+ bc-(b^2+c^2- bc)) = 32 bc (b+c )(a^2- (b-c )^2) = 32 6 9 (6+9 )(57- (6-9 )^2)=20.

$In △ ABC, if A is an acute angle, b = 6, c = 9 and sin A = \frac{2 14}{9}, then 3 a (cos B + cos C) =$

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