In $\triangle A B C$, if $a=13, b=14$ and $\cos \frac{C}{2}=\frac{3}{\sqrt{13}}$, then $2 r_1=$
In $\triangle A B C$, if $a=13, b=14$ and $\cos \frac{C}{2}=\frac{3}{\sqrt{13}}$, then $2 r_1=$
- 2 s
- $\Delta$
- s
- $2 \Delta$
Solution
$\begin{aligned} & \text { } a=13, b=14, \cos \frac{C}{2}=\frac{3}{\sqrt{13}} \\ & s=\frac{a+b+c}{2} \Rightarrow s=\frac{27+c}{2}\end{aligned}$
$\begin{aligned} & \cos \frac{C}{2}=\frac{3}{\sqrt{3}} \Rightarrow \sqrt{\frac{s(s-c)}{a b}}=\frac{3}{\sqrt{3}} \\ & \Rightarrow s(s-c)=126 \Rightarrow\left(\frac{27+c}{2}\right)\left(\frac{27-c}{2}\right)=126 \\ & \Rightarrow 729-c^2=504 \Rightarrow c=15 \text { and } s=21\end{aligned}$
$\begin{aligned} & \cos \frac{A}{2}=\sqrt{\frac{s(s-a)}{b c}}=\sqrt{\frac{21 \times 8}{14 \times 15}} \\ & \cos \frac{B}{2}=\sqrt{\frac{s(s-b)}{a c}}=\sqrt{\frac{21 \times 7}{13 \times 15}} \\ & 2 r_1=\frac{2 a \cos \frac{B}{2} \cos \frac{C}{2}}{\cos \frac{A}{2}}=\frac{2 \times 13 \times \sqrt{\frac{21 \times 7}{13 \times 15}} \times \frac{3}{\sqrt{13}}}{\sqrt{\frac{21 \times 8}{14 \times 15}}}=21=s\end{aligned}$
Asked in: AP EAMCET 2024 (21 May Shift 2)
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