Q67 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Mathematics

In a triangle $\mathrm{ABC}$, if $a=7, c=11, \cos \mathrm{A}=\frac{17}{22}, \cos \mathrm{C}=\frac{1}{14}$ then $b \tan \frac{\mathrm{B}}{2} \tan \frac{\mathrm{C}-\mathrm{A}}{2}=$

  • A
    18
  • B
    14
  • Correct answer
    C
    2
  • D
    9

Solution

Given $a=7, c=11, \cos \mathrm{A}=\frac{17}{22}, \cos \mathrm{C}=\frac{1}{14}$ Apply, law of tangent $$ \tan \left(\frac{\mathrm{c}-\mathrm{A}}{2}\right)=\left(\frac{\mathrm{c}-a}{\mathrm{c}+a}\right) \cot \left(\frac{\mathrm{B}}{2}\right) $$ Now, $\cos \mathrm{A}=\frac{b^2+c^2-a^2}{2 b c}$ $$ \begin{aligned} & \frac{17}{22}=\frac{b^2+121-49}{2 b \times 11} \\ & b^2-17 b+72=0 \\ & b^2-9 b-8 b+72=0 \\ & b(b-9)-8(b-9)=0 \\ & (b-8)(b-9)=0 \\ & b=8,9 \end{aligned} $$ Put $b=9$ in the given expression, Take, $$ \begin{aligned} & b \tan \frac{\mathrm{B}}{2} \tan \frac{\mathrm{c}-\mathrm{A}}{2}=\mathrm{b} \tan \frac{\mathrm{B}}{2}\left(\frac{11-7}{11+7}\right) \cot \left(\frac{\mathrm{B}}{2}\right) \\ & =9 \tan \left(\frac{\mathrm{B}}{2}\right) \times \frac{4}{18} \times \frac{1}{\tan \left(\frac{\mathrm{B}}{2}\right)} \\ & b \tan \frac{\mathrm{B}}{2} \tan \left(\frac{\mathrm{c}-\mathrm{A}}{2}\right)=2 \end{aligned} $$ So, correct option is (c).

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

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