In the given figure, ABCD is a kite with AC and BD as diagonals and $∠ A B C$ = $30^{\circ}$ . Find $x$ .

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Solution

From the figure, the diagonal AC divides the kite into two isosceles triangles.

In $△ A B C$ , $A B = C B$ (Adjacent sides of kite ABCD )

Given, $∠ A B C$ = 30°.
Since, $A B = B C, ∠ B A C = ∠ B C A$
Then, $∠ B C A = ∠ B A C = x$

By, angle sum property of a triangle,
$∠ A B C + ∠ B C A + ∠ B A C$ = 180°
$\Rightarrow 30 ° + x + x = 180 °$
$\Rightarrow 30 ° + 2 x = 180 °$
$\Rightarrow 2 x = 180 ° - 30 ° = 150 °$
$\Rightarrow x = 75 °$

Hence, the value of $x$ is $75 ° .$

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In the given figure, ABCD is a kite with AC and BD as diagonals and $∠ A B C$ = $30^{\circ}$ . Find $x$ .

Let ABCD be a kite with AC and BD as diagonals and $∠ A B C$ = $30^{\circ}$ . Then $∠ A C B$ =.

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∠ O B C = 20^{o}

∠ O C D = 40^{o}

∠

ABCD

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