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Satyam Singh · Answer 1 · 2023-02-09T23:01:54+0000

As given in the question, ABCD is a square. The angles between the sides of the square are always 90°. The diagonal divides this angle exactly into two halves.

Therefore, the angle BCA is

$\angle BCA=(90^0)/(2)=45^0^{}_{}$

The total angle at the centre where the diagonals meet is 360°.

These diagonals cut it into four equal parts.

Therefore, angle AWD is

$\angle AWD=(360^0)/(4)=90^0$

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