In a triangle ABC,if a=3, b=7, c=8,then sin f | vedclass

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(D) Given $a=3, b=7, c=8$. Using the Law of Tangents,$	an \frac{C-A}{2} = \frac{c-a}{c+a} \cot \frac{B}{2}$. Substituting this into the expression: $

Vedclass · Accepted Answer

$\frac{5 \sqrt{3}}{22}$

Vedclass · Answer

(D) Given $a=3, b=7, c=8$. Using the Law of Tangents,$	an \frac{C-A}{2} = \frac{c-a}{c+a} \cot \frac{B}{2}$. Substituting this into the expression: $\sin \frac{B}{2} 	an \frac{C-A}{2} = \sin \frac{B}{2} \left( \frac{c-a}{c+a} ight) \cot \frac{B}{2} = \sin \frac{B}{2} \left( \frac{c-a}{c+a} ight) \frac{\cos \frac{B}{2}}{\sin \frac{B}{2}} = \left( \frac{c-a}{c+a} ight) \cos \frac{B}{2}$. Now,calculate $\cos B$ using the Law of Cosines: $\cos B = \frac{a^2+c^2-b^2}{2ac} = \frac{3^2+8^2-7^2}{2 	imes 3 	imes 8} = \frac{9+64-49}{48} = \frac{24}{48} = \frac{1}{2}$. Using the half-angle formula $\cos \frac{B}{2} = \sqrt{\frac{1+\cos B}{2}}$: $\cos \frac{B}{2} = \sqrt{\frac{1+1/2}{2}} = \sqrt{\frac{3/2}{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$. Finally,substitute the values: $\left( \frac{8-3}{8+3} ight) \cos \frac{B}{2} = \frac{5}{11} 	imes \frac{\sqrt{3}}{2} = \frac{5 \sqrt{3}}{22}$. Thus,option $(d)$ is correct.

In a triangle $ABC$,if $a=3, b=7, c=8$,then $\sin \frac{B}{2} \tan \frac{C-A}{2} =$

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