In the given kite, $y$ equals

37^{\circ}

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28^{\circ}

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20^{\circ}

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40^{\circ}

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Solution

The correct option is A $28^{\circ}$
Given, ABCD is a kite,
Let the diagonals are meeting at O.
In $△ A B D$ and $△ B C D$
$A B = B C$ (Given)
$A D = C D$ (Given)
$B D = B D$ (Common)
Thus, $△ A B D ≅ △ C B D$ (SSS rule)
Thus, $∠ D B A = ∠ C B A$ (by CPCT) (1)
Now, In $△ O A B$ and $△ O C B$
$∠ O B A = ∠ O B C$ (From 1)
$O B = O B$ (Common)
$A B = B C$ (Given)
$△ O A B ≅ △ O C B$ (SAS rule)
Thus, $∠ A O B = ∠ C O B = \frac{180^{o}}{2} = 90^{\circ}$ (Equal angles by CPCT)
Now, in $△ O A B$
$∠ O A B + ∠ O B A + ∠ O A B = 180^{o}$
$37^{o} + 2 y - 3^{o} + 90^{o} = 180^{o}$
$2 y = 56^{o}$
$y = 28^{\circ}$

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