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If the radius of the incircle of a triangle with sides $5 k, 6 k$ and $5 k$ is 6 , then the largest angle of that triangle is
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Verified Answer
The correct answer is:
$\tan ^{-1}\left(\frac{24}{7}\right)$
Given, sides of $\triangle A B C$ is $5 k, 5 k$ and $6 k$ respectively,
$A B C$ is an isosceles triangle,
$\begin{aligned}
\Rightarrow \quad A D & =\sqrt{25 k^2-9 k^2}=4 k \Rightarrow \tan \theta=\frac{3}{4} \\
\angle B A C & =2 \theta \Rightarrow \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta} \\
\Rightarrow \quad \tan 2 \theta & =\frac{2 \times \frac{3}{4}}{1-\frac{9}{16}}=\frac{24}{7} \Rightarrow 2 \theta=\tan ^{-1}\left(\frac{24}{7}\right)
\end{aligned}$
$\therefore$ Largest angle is $\tan ^{-1}\left(\frac{24}{7}\right)$.
$A B C$ is an isosceles triangle,
$\begin{aligned}
\Rightarrow \quad A D & =\sqrt{25 k^2-9 k^2}=4 k \Rightarrow \tan \theta=\frac{3}{4} \\
\angle B A C & =2 \theta \Rightarrow \tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta} \\
\Rightarrow \quad \tan 2 \theta & =\frac{2 \times \frac{3}{4}}{1-\frac{9}{16}}=\frac{24}{7} \Rightarrow 2 \theta=\tan ^{-1}\left(\frac{24}{7}\right)
\end{aligned}$
$\therefore$ Largest angle is $\tan ^{-1}\left(\frac{24}{7}\right)$.
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