In a triangle ABC,if r: R: r_2 = 1: 3: 7,then | vedclass

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(D) Given $r: R: r_2 = 1: 3: 7$. Let $r = k, R = 3k, r_2 = 7k$. We know that $r_2 - r = 4R \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2} - 4R \si

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$2$

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(D) Given $r: R: r_2 = 1: 3: 7$. Let $r = k, R = 3k, r_2 = 7k$. We know that $r_2 - r = 4R \sin \frac{B}{2} \cos \frac{A}{2} \cos \frac{C}{2} - 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = 4R \sin \frac{B}{2} \cos \left( \frac{A+C}{2} \right)$. Substituting the values: $7k - k = 4(3k) \sin \frac{B}{2} \cos \left( \frac{\pi - B}{2} \right)$. $6k = 12k \sin \frac{B}{2} \sin \frac{B}{2} = 12k \sin^2 \frac{B}{2}$. $\sin^2 \frac{B}{2} = \frac{6k}{12k} = \frac{1}{2}$. Since $\sin^2 \frac{B}{2} = \frac{1 - \cos B}{2}$,we have $\frac{1 - \cos B}{2} = \frac{1}{2}$,which implies $\cos B = 0$,so $B = 90^{\circ}$. Now,$\sin(A+C) + \sin B = \sin(\pi - B) + \sin B = \sin B + \sin B = 2 \sin 90^{\circ} = 2(1) = 2$.

In a $\triangle ABC$,if $r: R: r_2 = 1: 3: 7$,then $\sin(A+C) + \sin B = $

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