In a triangle ABC,2A + C = 300^{circ}. If the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $2A + C = 300^{\circ}$ and $A + B + C = 180^{\circ}$.Subtracting the equations: $(2A + C) - (A + B + C) = 300^{\circ} - 180^{\circ} \implies

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(B) Given $2A + C = 300^{\circ}$ and $A + B + C = 180^{\circ}$.Subtracting the equations: $(2A + C) - (A + B + C) = 300^{\circ} - 180^{\circ} \implies A - B = 120^{\circ}$.Also,$R = 8r$. The formula for the inradius is $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.Substituting $R = 8r$,we get $r = 4(8r) \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \implies 32 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 1$.Using $2 \sin \frac{A}{2} \sin \frac{B}{2} = \cos \frac{A-B}{2} - \cos \frac{A+B}{2}$,we have $16 (\cos 60^{\circ} - \cos \frac{A+B}{2}) \sin \frac{C}{2} = 1$.Since $A+B = 180^{\circ} - C$,$\cos \frac{A+B}{2} = \sin \frac{C}{2}$.$16 (\frac{1}{2} - \sin \frac{C}{2}) \sin \frac{C}{2} = 1 \implies 8 \sin \frac{C}{2} - 16 \sin^2 \frac{C}{2} = 1$.$16 \sin^2 \frac{C}{2} - 8 \sin \frac{C}{2} + 1 = 0 \implies (4 \sin \frac{C}{2} - 1)^2 = 0$.Therefore,$\sin \frac{C}{2} = \frac{1}{4}$.

List-$I$	List-$II$
$(A)$ $rr_2 = r_1r_3$	$(I)$ $\angle A = 90^{\circ}$
$(B)$ $r_1 + r_2 = r_3 - r$	$(II)$ $b^2 = c^2 + a^2$
$(C)$ $r_1 = r + 2R$	$(III)$ $\angle C = 90^{\circ}$
	$(IV)$ $\angle B = 120^{\circ}$

List-$I$	List-$II$
$A. \tan \frac{A}{2} = \frac{r}{s-a}$	$I. (AI) \left( \frac{\sqrt{(s-b)(s-c)}}{bc} \right)$
$B. r$	$II. R^2$
$C. (SI)^2 + 2Rr$	$III. (4R + r + \sqrt{2}s)(4R + r - \sqrt{2}s)$
$D. r_1^2 + r_2^2 + r_3^2$	$IV. \frac{Rr}{S}$
	$V. \frac{(s-b)(s-c)}{\Delta}$

In a $\triangle ABC$,$2A + C = 300^{\circ}$. If the circumradius $R$ of the $\triangle ABC$ is eight times its inradius $r$,then $\sin \frac{C}{2} = $

Similar Questions

The perimeter of $\triangle ABC$ is $36 \text{ cm}$ and its inradius is $8 \text{ cm}$. Then,the area of the triangle is (in $\text{ cm}^2$)

In an equilateral triangle $ABC$, the ratio $r : R : r_1$ is

$PQR$ is an isosceles triangle with $PQ=PR$. If the radius of the circumcircle of $\triangle PQR$ is equal to the length of $PQ$,then $\angle P=$ (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module