In the following figure,in Delta ABC,angle AB | vedclass

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(N/A) Given that $\angle ABC = 70^{\circ}$ and $\angle ACB = 60^{\circ}$.Since $DBC$ is a straight line,$\angle ABC$ and $\angle ABD$ form a linear pa

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(N/A) Given that $\angle ABC = 70^{\circ}$ and $\angle ACB = 60^{\circ}$.
Since $DBC$ is a straight line,$\angle ABC$ and $\angle ABD$ form a linear pair.
Therefore,$\angle ABD + \angle ABC = 180^{\circ}$.
$\angle ABD + 70^{\circ} = 180^{\circ} \implies \angle ABD = 180^{\circ} - 70^{\circ} = 110^{\circ}$.
Similarly,$\angle ACB$ and $\angle ACE$ form a linear pair on the straight line $BCE$.
Therefore,$\angle ACE + \angle ACB = 180^{\circ}$.
$\angle ACE + 60^{\circ} = 180^{\circ} \implies \angle ACE = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Thus,$\angle ABD = 110^{\circ}$ and $\angle ACE = 120^{\circ}$.

In the following figure,in $\Delta ABC$,$\angle ABC = 70^{\circ}$ and $\angle ACB = 60^{\circ}$. Find $\angle ABD$ and $\angle ACE$.

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