In an isosceles right-angled triangle,a strai | vedclass

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(B) Let the vertices be $A, B, C$ such that $\angle A = 90^{\circ}$ and $AB = AC = a$. Let $D$ be the midpoint of $AC$. Then $AD = DC = \frac{a}{2}$.I

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$2$ and $3$

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(B) Let the vertices be $A, B, C$ such that $\angle A = 90^{\circ}$ and $AB = AC = a$. Let $D$ be the midpoint of $AC$. Then $AD = DC = \frac{a}{2}$.In $	riangle ADB$,$\angle DAB = 90^{\circ}$. Let $\angle ABD = \alpha$. Then $\cot \alpha = \frac{AB}{AD} = \frac{a}{a/2} = 2$.Let $\angle DBC = \beta$. Since $\angle ABC = 45^{\circ}$,we have $\alpha + \beta = 45^{\circ}$.Using the formula $\cot(\alpha + \beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot \beta}$,we get $\cot 45^{\circ} = 1$.$\frac{2 \cot \beta - 1}{2 + \cot \beta} = 1$ $\Rightarrow 2 \cot \beta - 1 = 2 + \cot \beta$ $\Rightarrow \cot \beta = 3$.Thus,the cotangents of the two angles are $2$ and $3$.

In an isosceles right-angled triangle,a straight line is drawn from the midpoint 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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