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In an isosceles right angled triangle, a straight line is drawn from the mid point of one of the equal sides to the opposite vertex. Then a pair of possible values of the cotangents of the two angles so formed at that vertex are
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The correct answer is:
2 and 3
$A B=A C=a$
$\begin{aligned}
& A D=D C=\frac{a}{2} \\
& \angle C=\angle A B C=45^{\circ}
\end{aligned}$
In $\triangle A D B$,
$\begin{aligned}
& \cot (D B A)=\frac{A B}{A D}=\frac{a}{a / 2}=2 \\
& \therefore \quad \cot \alpha=2 \\
& \because \quad \alpha+\beta=45^{\circ} \\
& \Rightarrow \quad \cot (\alpha+\beta)=\cot 45^{\circ} \\
& \Rightarrow \quad \frac{\cot \alpha \cdot \cot \beta-1}{\cot \beta+\cot \alpha}=1
\end{aligned}$
$\begin{aligned} & \Rightarrow \cot \beta \cdot \cot \alpha-1=\cot \beta+\cot \alpha \\ & \Rightarrow \quad 2 \cot \beta-1=2+\cot \beta \\ & \Rightarrow \cot \beta=3 \\ & \therefore \quad \cot \alpha=2, \cot \beta=3 .\end{aligned}$
$\begin{aligned}
& A D=D C=\frac{a}{2} \\
& \angle C=\angle A B C=45^{\circ}
\end{aligned}$
In $\triangle A D B$,
$\begin{aligned}
& \cot (D B A)=\frac{A B}{A D}=\frac{a}{a / 2}=2 \\
& \therefore \quad \cot \alpha=2 \\
& \because \quad \alpha+\beta=45^{\circ} \\
& \Rightarrow \quad \cot (\alpha+\beta)=\cot 45^{\circ} \\
& \Rightarrow \quad \frac{\cot \alpha \cdot \cot \beta-1}{\cot \beta+\cot \alpha}=1
\end{aligned}$
$\begin{aligned} & \Rightarrow \cot \beta \cdot \cot \alpha-1=\cot \beta+\cot \alpha \\ & \Rightarrow \quad 2 \cot \beta-1=2+\cot \beta \\ & \Rightarrow \cot \beta=3 \\ & \therefore \quad \cot \alpha=2, \cot \beta=3 .\end{aligned}$
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