Match the items of List-I with those of List-II (Here $\Delta$ denotes the area of $\triangle A B C$.) Then…
Match the items of List-I with those of List-II (Here $\Delta$ denotes the area of $\triangle A B C$.)
Then the correct match is
- (A) - (vi), (B) - (i), (C) - (ii), (D) - (iii)
- (A) - (ii), (B) - (i), (C) - (v), (D) - (iii)
- (A) - (ii), (B) - (vi), (C) - (v), (D) - (i)
- (A) - (vi), (B) - (ii), (C) - (i), (D) - (iv)
Solution
$\cot A=\frac{b^2+c^2-a^2}{4 \Delta}$
$\cot B=\frac{c^2+a^2-b^2}{4 \Delta} \Rightarrow \cot C=\frac{a^2+b^2-c^2}{4 \Delta}$
$\sum \cot A=\cos A+\cot B+\cot C=\frac{a^2+b^2+c^2}{4 \Delta}$
$\cot \frac{A}{2}=\frac{s(s-a)}{\Delta}, \cot \frac{B}{2}=\frac{s(s-b)}{\Delta}$ and $\cot \frac{C}{2}=\frac{s(s-c)}{\Delta}$
$\sum \cot \frac{A}{2}=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}$
$=\frac{s}{\Delta}(s-a+s-b+s-c)=\frac{s}{\Delta}(3 s-2 s)$
$=\frac{s^2}{\Delta}=\frac{(a+b+c)^2}{4 \Delta}$
$\tan A: \tan B: \tan C=1: 2: 3$
Let $\tan A=k, \tan B=2 k, \tan C=3 k$
$\Rightarrow \sin A=\frac{k}{\sqrt{1+k^2}}, \sin B=\frac{2 k}{\sqrt{1+4 k^2}}, \sin C=\frac{3 k}{\sqrt{1+4 k^2}}$
$\because A+B+C=\pi \therefore \sum \tan A=\pi \tan A \therefore 6 k=6 k^3 \Rightarrow k=1$
$\therefore \sin A=\frac{1}{\sqrt{2}}, \sin B=\frac{2}{\sqrt{5}}, \sin C=\frac{3}{\sqrt{10}}$
$\therefore \sin A: \sin B: \sin C=\sqrt{5}: 2 \sqrt{2}: 3$
$\cot \frac{A}{2}: \cot \frac{B}{2}: \cot \frac{C}{2}=3: 7: 9$
$\Rightarrow \frac{s(s-a)}{\Delta}: \frac{s(s-b)}{\Delta}: \frac{s(s-c)}{\Delta}=3: 7: 9$
$\Rightarrow s-a: s-b: s-c=3: 7: 9$
$\Rightarrow s-a=3 k, s-b=7 k, s-c=9 k$
$\Rightarrow b+c-a=6 k, a+c-b=14 k, a+b-c=18 k$
Adding first two equations
$c=10 k, a=16 k, b=12 k \Rightarrow a: b: c=8: 6: 5$
Asked in: AP EAMCET 2024 (20 May Shift 1)
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