In a triangle ABC,if b=10,a cos^2 frac{C}{2} | vedclass

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(D) Given $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = 15$. Using $\cos^2 	heta = \frac{1+\cos 2	heta}{2}$,we get $\frac{a(1+\cos C)}{2} + \frac{c

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$\frac{5}{\sqrt{3}}$

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(D) Given $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = 15$. Using $\cos^2 	heta = \frac{1+\cos 2	heta}{2}$,we get $\frac{a(1+\cos C)}{2} + \frac{c(1+\cos A)}{2} = 15$. $\Rightarrow (a+c) + (a \cos C + c \cos A) = 30$. Since $a \cos C + c \cos A = b$,we have $a+c+b = 30$. Given $b=10$,so $a+c = 20$. The semi-perimeter $s = \frac{a+b+c}{2} = \frac{30}{2} = 15$. Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = 15\sqrt{3}$. $15\sqrt{3} = \sqrt{15(15-a)(15-10)(15-c)} = \sqrt{75(15-a)(15-c)}$. Squaring both sides: $225 	imes 3 = 75(15-a)(15-c) \Rightarrow 9 = (15-a)(15-c) = 225 - 15(a+c) + ac$. $9 = 225 - 15(20) + ac$ $\Rightarrow 9 = 225 - 300 + ac$ $\Rightarrow ac = 84$. Area $\Delta = \frac{1}{2}ac \sin B = 15\sqrt{3}$ $\Rightarrow \frac{1}{2}(84) \sin B = 15\sqrt{3}$ $\Rightarrow 42 \sin B = 15\sqrt{3}$ $\Rightarrow \sin B = \frac{5\sqrt{3}}{14}$. Then $\cos^2 B = 1 - \sin^2 B = 1 - \frac{75}{196} = \frac{121}{196} \Rightarrow \cos B = \frac{11}{14}$. Finally,$\cot \frac{B}{2} = \sqrt{\frac{1+\cos B}{1-\cos B}} = \sqrt{\frac{1 + 11/14}{1 - 11/14}} = \sqrt{\frac{25/14}{3/14}} = \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}}$.

In a $\triangle ABC$,if $b=10$,$a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = 15$,and the area of the triangle is $15\sqrt{3}$ sq. units,then $\cot \frac{B}{2} =$

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