Q74 / 160 · TS EAMCET 2004 · Mathematics

In a $\triangle A B C$, the correct formulae among the following are

  • A
    only I, II
  • B
    only I,II
  • Correct answer
    C
    only I,III
  • D
    I,II,III

Solution

I. Since, $4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ $$ \begin{aligned} & =\frac{4 a b c}{4 \Delta} \sqrt{\frac{(s-b)(s-c)}{b c}} \sqrt{\frac{(s-a)(s-c)}{a c}} \\ & =\frac{a b c}{\Delta} \sqrt{\frac{(s-a)(s-b)}{a b}} \\ & =\frac{a b c}{4 \Delta} \frac{\Delta^2}{s \cdot a b c} \\ & =\frac{\Delta}{s}=r \end{aligned} $$ Hence, $r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$ II. Since, $$ \begin{aligned} & (s-a) \tan A / 2=(s-a) \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \\ & =\sqrt{\frac{s(s-a)(s-b)(s-c)}{s^2}} \\ & =\frac{\Delta}{s}=r \end{aligned} $$ Hence, $r_1 \neq(s-a) \tan \frac{A}{2}$ III. Since, $\frac{\Delta}{s-c}=\sqrt{\frac{s(s-a)(s-b)}{(s-c)}}$ $$ =s \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=s \tan \frac{c}{2}=r_3 $$ Hence $r_3=\frac{\Delta}{s-c}$ Therefore I and III statements are true but II is a false statement.

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

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