In triangle ABC,if A=45^{circ},C=75^{circ} an | vedclass

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Question

(A) In $	riangle ABC$,the sum of angles is $180^{\circ}$. Given $A=45^{\circ}$ and $C=75^{\circ}$,we have $B = 180^{\circ} - (45^{\circ} + 75^{\circ}

Vedclass · Accepted Answer

$\frac{\sqrt{3}+1}{1+\sqrt{2}+\sqrt{3}}$

Vedclass · Answer

(A) In $	riangle ABC$,the sum of angles is $180^{\circ}$. Given $A=45^{\circ}$ and $C=75^{\circ}$,we have $B = 180^{\circ} - (45^{\circ} + 75^{\circ}) = 60^{\circ}$.Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R = 2\sqrt{2}$.$a = 2R \sin A = 2\sqrt{2} \sin 45^{\circ} = 2\sqrt{2} 	imes \frac{1}{\sqrt{2}} = 2$.$b = 2R \sin B = 2\sqrt{2} \sin 60^{\circ} = 2\sqrt{2} 	imes \frac{\sqrt{3}}{2} = \sqrt{6}$.$c = 2R \sin C = 2\sqrt{2} \sin 75^{\circ} = 2\sqrt{2} 	imes \sin(45^{\circ}+30^{\circ}) = 2\sqrt{2} 	imes (\frac{\sqrt{3}+1}{2\sqrt{2}}) = \sqrt{3}+1$.The semi-perimeter $s = \frac{a+b+c}{2} = \frac{2+\sqrt{6}+\sqrt{3}+1}{2} = \frac{3+\sqrt{3}+\sqrt{6}}{2}$.The inradius $r$ is given by $r = 4R \sin(\frac{A}{2}) \sin(\frac{B}{2}) \sin(\frac{C}{2})$ or $r = \frac{abc}{4RS}$.Using $r = (s-a) 	an(\frac{A}{2})$,or simply calculating from the formula $r = \frac{abc}{4RS}$:$r = \frac{2 	imes \sqrt{6} 	imes (\sqrt{3}+1)}{4 	imes \sqrt{2} 	imes \frac{3+\sqrt{3}+\sqrt{6}}{2}} = \frac{2\sqrt{6}(\sqrt{3}+1)}{2\sqrt{2}(3+\sqrt{3}+\sqrt{6})} = \frac{\sqrt{3}(\sqrt{3}+1)}{3+\sqrt{3}+\sqrt{6}} = \frac{\sqrt{3}+1}{\sqrt{3}+1+\sqrt{2}}$.Thus,the correct option is $A$.

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 $a : b : c =$	$(iv)$ $12 : 5 : 13$
	$(v)$ $\sqrt{5} : 2\sqrt{2} : 3$
	$(vi)$ $4\Delta$

In triangle $ABC$,if $A=45^{\circ}$,$C=75^{\circ}$ and $R=\sqrt{2}$,then $r=$

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