$\left(1\right)\because \angle B=50^{\circ}$，$\angle C=70^{\circ}$，
$\therefore \angle BAC=180^{\circ}-\angle B-\angle C=180^{\circ}-50^{\circ}-70^{\circ}=60^{\circ}$.
$\because AD$是$\triangle ABC$的角平分线，
$\therefore \angle BAD=\frac{1}{2}\angle BAC=\frac{1}{2}\times 60^{\circ}=30^{\circ}$.
$\because DE\bot AB$，
$\therefore \angle DEA=90^{\circ}$，
$\therefore \angle EDA=180^{\circ}-\angle BAD-\angle DEA=180^{\circ}-30^{\circ}-90^{\circ}=60^{\circ}$；
$(2)$如图，过$D$作$DF\bot AC$于点$F$，

$\because AD$是$\triangle ABC$的角平分线，$DE\bot AB$，
$\therefore DF=DE=3$，
又$\because AB=10$，$AC=8$，
$\therefore S_{\triangle ABC}=\frac{1}{2}\times AB\times DE+\frac{1}{2}\times AC\times DF=\frac{1}{2}\times 10\times 3+\frac{1}{2}\times 8\times 3=27$.

$\left(1\right)\because \angle B=50^{\circ}$，$\angle C=70^{\circ}$，
$\therefore \angle BAC=180^{\circ}-\angle B-\angle C=180^{\circ}-50^{\circ}-70^{\circ}=60^{\circ}$.
$\because AD$是$\triangle ABC$的角平分线，
$\therefore \angle BAD=\frac{1}{2}\angle BAC=\frac{1}{2}\times 60^{\circ}=30^{\circ}$.
$\because DE\bot AB$，
$\therefore \angle DEA=90^{\circ}$，
$\therefore \angle EDA=180^{\circ}-\angle BAD-\angle DEA=180^{\circ}-30^{\circ}-90^{\circ}=60^{\circ}$；
$(2)$如图，过$D$作$DF\bot AC$于点$F$，

$\because AD$是$\triangle ABC$的角平分线，$DE\bot AB$，
$\therefore DF=DE=3$，
又$\because AB=10$，$AC=8$，
$\therefore S_{\triangle ABC}=\frac{1}{2}\times AB\times DE+\frac{1}{2}\times AC\times DF=\frac{1}{2}\times 10\times 3+\frac{1}{2}\times 8\times 3=27$.