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 $$0$$


Answer & Solution

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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