In a triangle ABC,if A-B=120^{circ} and R=8r, | vedclass

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(C) Given $A-B=120^{\circ}$ and $R=8r$. We know that $r = 4R \sin(A/2) \sin(B/2) \sin(C/2)$. Since $R=8r$,we have $r/R = 1/8$,so $4 \sin(A/2) \sin(B/2

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) Given $A-B=120^{\circ}$ and $R=8r$. We know that $r = 4R \sin(A/2) \sin(B/2) \sin(C/2)$. Since $R=8r$,we have $r/R = 1/8$,so $4 \sin(A/2) \sin(B/2) \sin(C/2) = 1/8$,which implies $\sin(A/2) \sin(B/2) \sin(C/2) = 1/32$. Using $2 \sin(A/2) \sin(B/2) = \cos((A-B)/2) - \cos((A+B)/2) = \cos(60^{\circ}) - \cos(90^{\circ}-C/2) = 1/2 - \sin(C/2)$. Substituting this,$(1/2 - \sin(C/2)) \sin(C/2) = 1/16$,so $\sin^2(C/2) - 1/2 \sin(C/2) + 1/16 = 0$. This is $(\sin(C/2) - 1/4)^2 = 0$,so $\sin(C/2) = 1/4$. Then $\cos C = 1 - 2 \sin^2(C/2) = 1 - 2(1/16) = 1 - 1/8 = 7/8$. Finally,$\frac{1+\cos C}{1-\cos C} = \frac{1+7/8}{1-7/8} = \frac{15/8}{1/8} = 15$.

List-$I$	List-$II$
$(A) \ r_1 r_2 \sqrt{\frac{4R-r_1-r_2}{r_1+r_2}}$	$1. \ b$
$(B) \ \frac{r_2(r_3+r_1)}{\sqrt{r_1r_2+r_2r_3+r_3r_1}}$	$2. \ a^2, b^2, c^2 \text{ are in } AP$
$(C) \ \frac{a}{c} = \frac{\sin(A-B)}{\sin(B-C)}$	$3. \ \Delta$
$(D) \ bc \cos^2 \frac{A}{2}$	$4. \ R r_1 r_2 r_3$
	$5. \ s(s-a)$

In a $\triangle ABC$,if $A-B=120^{\circ}$ and $R=8r$,then find the value of $\frac{1+\cos C}{1-\cos C}$.

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