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

• A

  $\frac{15 \sqrt{3}}{22 \sqrt{7}}$
• B

  $\frac{5 \sqrt{2}}{11 \sqrt{7}}$
• C

  $\frac{5 \sqrt{3}}{11}$
• Correct answer
  D

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

Solution

Given $\mathrm{a}=3, \mathrm{~b}=7, \mathrm{c}=8$ Now, $\sin \left(\frac{\mathrm{B}}{2}\right) \tan \left(\frac{\mathrm{C}-\mathrm{A}}{2}\right)=\sin \frac{\mathrm{B}}{2}\left(\frac{\mathrm{c}-\mathrm{a}}{\mathrm{c}+\mathrm{a}}\right) \cot \frac{\mathrm{B}}{2}$ \{By Law of tangent\}
Take, $\cos ^2 \frac{\mathrm{B}}{2}=\frac{1+\cos \mathrm{B}}{2}$ So, $\cos \mathrm{B}=\frac{\mathrm{a}^2+\mathrm{c}^2-\mathrm{b}^2}{2 \mathrm{ac}}=\frac{9+64-49}{2 \times 3 \times 8}=\frac{24}{6 \times 8}=\frac{1}{2}$ Put value of $\cos B$ in the value $\cos ^2\left(\frac{B}{2}\right)$ $$ \begin{aligned} & \cos ^2 \frac{B}{2}=\frac{1+\frac{1}{2}}{2}=\frac{3}{4} \\ & \cos \left(\frac{B}{2}\right)=\frac{\sqrt{3}}{2} \end{aligned} $$ From equation (i), $$ \sin \frac{\mathrm{B}}{2} \tan \frac{\mathrm{C}-\mathrm{A}}{2}=\frac{5}{11} \times \frac{\sqrt{3}}{2}=\frac{5 \sqrt{3}}{22} $$ So, option (d) is correct.

Hence, the correct answer is option: 'D'.