ABCD is a rhombus such that angle ACB = 40^{c | vedclass

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(D) Given that $ABCD$ is a rhombus and $\angle ACB = 40^{\circ}$.We know that the diagonals of a rhombus bisect each other at right angles $(90^{\circ

Vedclass · Accepted Answer

$50$

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(D) Given that $ABCD$ is a rhombus and $\angle ACB = 40^{\circ}$.We know that the diagonals of a rhombus bisect each other at right angles $(90^{\circ})$.Let the diagonals intersect at point $O$. Thus,$	riangle BOC$ is a right-angled triangle with $\angle BOC = 90^{\circ}$.In $	riangle BOC$,the sum of angles is $180^{\circ}$:$\angle OBC + \angle BOC + \angle BCO = 180^{\circ}$$\angle OBC + 90^{\circ} + 40^{\circ} = 180^{\circ}$$\angle OBC = 180^{\circ} - 130^{\circ} = 50^{\circ}$.Since $AD \parallel BC$ (opposite sides of a rhombus are parallel),the alternate interior angles are equal:$\angle ADB = \angle DBC$.Since $\angle DBC = \angle OBC = 50^{\circ}$,we have $\angle ADB = 50^{\circ}$.

$ABCD$ is a rhombus such that $\angle ACB = 40^{\circ}$. Then $\angle ADB$ is (in $^{\circ}$)

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