Mathematics

In the figure given above, if $AC= AD= CD= BD$; then $\angle ABC$ is $30.5^{\circ}$If true then enter $1$ and if false then enter $0$

1

SOLUTION
In $\triangle$ ACD,
Since, AC = CD
$\angle CAD = \angle CDA = x$ (Angles opposite to equal sides are equal)
Sum of angles of the triangle,
$\angle ACD + \angle CAD + \angle CDA = 180$
$58 + 2x = 180$
$2x = 122$
$x = 61$
In $\triangle ABD$,
AD = BD thus, $\angle ABD = \angle DAB = y$
Also, $\angle ABD + \angle DAB = \angle ADC$ (Exterior angle is equal to sum of opposite interior angles)
$2\angle ABD = 61$
$\angle ABD = 30.5^{\circ}$

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