In the given figure, find
$∠ D F E$

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Solution

$\begin{matrix} S u m o f a n g l e o f Δ D E F = 180^{0} ∴ ∠ D + ∠ D E F + ∠ D F E = 180^{0} 90^{0} + 45^{0} + ∠ D F E = 180^{0} ∠ D F E = 180^{o} - 90^{0} - 45^{0} = 45^{0} \end{matrix}$

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