Match the items of List-I with those of List- | vedclass

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(B) $\sum \cot A = \cot A + \cot B + \cot C = \frac{b^2+c^2-a^2}{4\Delta} + \frac{c^2+a^2-b^2}{4\Delta} + \frac{a^2+b^2-c^2}{4\Delta} = \frac{a^2+b^2+

Vedclass · Accepted Answer

$(A)$ - $(ii)$,$(B)$ - $(i)$,$(C)$ - $(v)$,$(D)$ - $(iii)$

Vedclass · Answer

(B) $\sum \cot A = \cot A + \cot B + \cot C = \frac{b^2+c^2-a^2}{4\Delta} + \frac{c^2+a^2-b^2}{4\Delta} + \frac{a^2+b^2-c^2}{4\Delta} = \frac{a^2+b^2+c^2}{4\Delta}$. Thus,$(A)$ matches $(ii)$.$(B)$ $\sum \cot \frac{A}{2} = \frac{s(s-a)}{\Delta} + \frac{s(s-b)}{\Delta} + \frac{s(s-c)}{\Delta} = \frac{s}{\Delta}(3s - (a+b+c)) = \frac{s}{\Delta}(3s - 2s) = \frac{s^2}{\Delta} = \frac{(a+b+c)^2}{4\Delta}$. Thus,$(B)$ matches $(i)$.$(C)$ Given $	an A : 	an B : 	an C = 1 : 2 : 3$. Let $	an A = k, 	an B = 2k, 	an C = 3k$. Since $A+B+C = \pi$,$	an A + 	an B + 	an C = 	an A 	an B 	an C \Rightarrow 6k = 6k^3 \Rightarrow k=1$. So $	an A = 1, 	an B = 2, 	an C = 3$. Then $\sin A = \frac{1}{\sqrt{2}}, \sin B = \frac{2}{\sqrt{5}}, \sin C = \frac{3}{\sqrt{10}}$. Ratio $\sin A : \sin B : \sin C = \frac{1}{\sqrt{2}} : \frac{2}{\sqrt{5}} : \frac{3}{\sqrt{10}} = \sqrt{5} : 2\sqrt{2} : 3$. Thus,$(C)$ matches $(v)$.$(D)$ Given $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$. Since $\cot \frac{A}{2} = \frac{s(s-a)}{\Delta}$,we have $(s-a) : (s-b) : (s-c) = 3 : 7 : 9$. Let $s-a=3k, s-b=7k, s-c=9k$. Adding gives $3s - (a+b+c) = 19k \Rightarrow s = 19k$. Then $a = 16k, b = 12k, c = 10k$. Ratio $a : b : c = 16 : 12 : 10 = 8 : 6 : 5$. Thus,$(D)$ matches $(iii)$.

List-$I$	List-$II$
$(A)$ $\sum \cot A$	$(i)$ $\frac{(a+b+c)^2}{4\Delta}$
$(B)$ $\sum \cot \frac{A}{2}$	$(ii)$ $\frac{a^2+b^2+c^2}{4\Delta}$
$(C)$ If $\tan A : \tan B : \tan C = 1 : 2 : 3$,then $\sin A : \sin B : \sin C =$	$(iii)$ $8 : 6 : 5$
$(D)$ If $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$,then $a : b : c =$	$(iv)$ $12 : 5 : 13$
	$(v)$ $\sqrt{5} : 2\sqrt{2} : 3$
	$(vi)$ $4\Delta$

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