In triangle ABC,if A is an acute angle,b=6, c | vedclass

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(B) Given $b=6, c=9$ and $\sin A=\frac{2 \sqrt{14}}{9}$.Since $A$ is an acute angle,$\cos A = \sqrt{1-\sin^2 A} = \sqrt{1-\frac{56}{81}} = \frac{5}{9}

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) Given $b=6, c=9$ and $\sin A=\frac{2 \sqrt{14}}{9}$.Since $A$ is an acute angle,$\cos A = \sqrt{1-\sin^2 A} = \sqrt{1-\frac{56}{81}} = \frac{5}{9}$.Using the Law of Cosines,$\cos A = \frac{b^2+c^2-a^2}{2bc}$.$\frac{5}{9} = \frac{36+81-a^2}{2(6)(9)}$ $\Rightarrow \frac{5}{9} = \frac{117-a^2}{108}$ $\Rightarrow 60 = 117-a^2$ $\Rightarrow a^2 = 57$.Now,$3a(\cos B+\cos C) = 3a\left(\frac{a^2+c^2-b^2}{2ac} + \frac{a^2+b^2-c^2}{2ab}ight)$.$= 3a\left(\frac{b(a^2+c^2-b^2) + c(a^2+b^2-c^2)}{2abc}ight) = \frac{3}{2bc} [a^2(b+c) + bc^2 - b^3 + b^2c - c^3]$.$= \frac{3}{2bc} [a^2(b+c) + bc(b+c) - (b+c)(b^2-bc+c^2)]$.$= \frac{3(b+c)}{2bc} [a^2 + bc - b^2 + bc - c^2] = \frac{3(b+c)}{2bc} [a^2 - (b-c)^2]$.Substituting the values: $\frac{3(6+9)}{2(6)(9)} [57 - (6-9)^2] = \frac{3(15)}{108} [57 - 9] = \frac{45}{108} 	imes 48 = \frac{45 	imes 4}{9} = 5 	imes 4 = 20$.

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

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