Q69 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
In a triangle ABC , if $r_1=6, r_2=9, r_3=18$, then $\cos \mathrm{A}=$
- A$\frac{5}{13}$
- Correct answerB$\frac{4}{5}$
- C$\frac{5}{7}$
- D$\frac{7}{25}$
Solution
$\frac{\Delta}{s-a}=6 \Rightarrow s-a=\frac{\Delta}{6}$ ...(i)
$\frac{\Delta}{s-b}=9 \Rightarrow s-b=\frac{\Delta}{9}$ ...(ii)
$\frac{\Delta}{s-c}=18 \Rightarrow s-c=\frac{\Delta}{18}$ ...(iii)
Adding (i), (ii) and (iii)
$\begin{aligned}
& 3 s-(a+b+c)=\frac{\Delta}{3} \Rightarrow s=\frac{\Delta}{3} ; \frac{\Delta}{3}-a=\frac{\Delta}{6} \Rightarrow a=\frac{3 \Delta}{18} \\
& \frac{\Delta}{3}-b=\frac{\Delta}{9} \Rightarrow b=\frac{4 \Delta}{18} ; \frac{\Delta}{3}-c=\frac{\Delta}{18} \Rightarrow c=\frac{5 \Delta}{18} \\
& a: b: c=3: 4: 5 \Rightarrow a=3 k, b=4 k, c=5 k \\
& \cos A=\frac{b^2+c^2-a^2}{2 b c}=\frac{(16+25-9) k^2}{(2 \times 4 \times 5) k^2} \Rightarrow \cos A=\frac{4}{5} .
\end{aligned}$
Hence, the correct answer is option: 'B'.