If the area of a triangle ABC is 4sqrt{5} tex | vedclass

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(C) Given: Area $\Delta = 4\sqrt{5}$, $b = 6$, and $	an \frac{B}{2} = \frac{\sqrt{5}}{4}$.Using the formula $	an \frac{B}{2} = \sqrt{\frac{(s-a)(s-c

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$3$

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(C) Given: Area $\Delta = 4\sqrt{5}$, $b = 6$, and $	an \frac{B}{2} = \frac{\sqrt{5}}{4}$.Using the formula $	an \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$, where $s = \frac{a+b+c}{2}$.We know $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$, so $(s-a)(s-c) = \frac{\Delta^2}{s(s-b)}$.Substituting this into the tangent formula: $	an^2 \frac{B}{2} = \frac{\Delta^2}{s^2(s-b)^2}$.Given $	an \frac{B}{2} = \frac{\sqrt{5}}{4}$, so $	an^2 \frac{B}{2} = \frac{5}{16}$.$\frac{5}{16} = \frac{(4\sqrt{5})^2}{s^2(s-6)^2} = \frac{80}{s^2(s-6)^2}$.$s^2(s-6)^2 = \frac{80 	imes 16}{5} = 256$.$s(s-6) = 16$ or $s(s-6) = -16$ (impossible).$s^2 - 6s - 16 = 0 \implies (s-8)(s+2) = 0$. Since $s > 0$, $s = 8$.$s = \frac{a+b+c}{2} = 8 \implies a+c+6 = 16 \implies a+c = 10$.Also, $\Delta = \frac{1}{2}ac \sin B = 4\sqrt{5}$.Using $	an \frac{B}{2} = \frac{\sqrt{5}}{4}$, $\cos B = \frac{1 - 	an^2(B/2)}{1 + 	an^2(B/2)} = \frac{1 - 5/16}{1 + 5/16} = \frac{11/16}{21/16} = \frac{11}{21}$.$\sin B = \sqrt{1 - (11/21)^2} = \frac{\sqrt{441-121}}{21} = \frac{\sqrt{320}}{21} = \frac{8\sqrt{5}}{21}$.$\frac{1}{2}ac \left(\frac{8\sqrt{5}}{21}ight) = 4\sqrt{5} \implies ac = 21$.We have $a+c = 10$ and $ac = 21$. The roots of $x^2 - 10x + 21 = 0$ are $x = 3, 7$.The sides are $3, 6, 7$. The smallest side is $3$.

If the area of a triangle $ABC$ is $4\sqrt{5} \text{ sq. units}$, the length of the side $CA$ is $6 \text{ units}$, and $\tan \frac{B}{2} = \frac{\sqrt{5}}{4}$, then the length of its smallest side is: (in $\text{ units}$)

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