$\widehat B:\widehat C=4:3\\↔\dfrac{\widehat B}{\widehat C}=\dfrac{4}{3}\\↔\widehat B=\dfrac{4\widehat C}{3}$

Xét $ΔABC$:

$\widehat A+\widehat B+\widehat C=180^\circ\\↔75^\circ+\dfrac{4\widehat C}{3}+\widehat C=180^\circ\\↔\dfrac{7\widehat C}{3}=105^\circ\\↔7\widehat C=315^\circ\\↔\widehat C=45^\circ\\→\widehat B=60^\circ$

Kẻ đường cao $AH$

Xét $ΔABH$ vuông tại $H$:

$\sin\widehat B=\dfrac{AH}{AB}$ hay $\sin 60^\circ=\dfrac{AH}{10}$

$↔\dfrac{\sqrt 3}{2}=\dfrac{AH}{10}$

$↔5\sqrt 3cm=AH$

Xét $ΔACH$ vuông tại $H$:

$\sin\widehat C=\dfrac{AH}{AC}$ hay $\sin 45^\circ=\dfrac{5\sqrt 3}{AC}$

$↔\dfrac{\sqrt 2}{2}=\dfrac{5\sqrt 3}{AC}$

$↔AC=5\sqrt 6cm$

Xét $ΔABH$ vuông tại $H$:

$\cos\widehat B=\dfrac{BH}{AB}$ hay $\cos 60^\circ=\dfrac{BH}{10}$

$↔\dfrac{1}{2}=\dfrac{BH}{10}\\↔BH=5cm$

Xét $ΔACH$ vuông tại $H$:

$\tan\widehat C=\dfrac{AH}{CH}$ hay $\tan 45^\circ=\dfrac{5\sqrt 3}{CH}$

$↔1=\dfrac{5\sqrt 3}{CH}\\↔5\sqrt 3cm=CH$

Ta có: $BC=BH+CH=5+5\sqrt 3$

$S_{ΔABC}=\dfrac{1}{2}.AH.BC=\dfrac{1}{2}.5\sqrt 3.(5+5\sqrt 3)=\dfrac{75+25\sqrt 3}{2}cm^2$

Vậy $AC=5\sqrt 6cm,BC=5+5\sqrt 3,S_{ΔABC}=\dfrac{75+25\sqrt 3}{2}cm^2$