Find the area of the right triangle with hypotenuse $8cm$ and one of the acute angle is ${57}^{\circ }$

Let $△ABC$ is the right angle triangle, right angled at $B$ & $\angle C=57°$

Now,

$\Rightarrow \sin 57°=\frac{AB}{AC}=\frac{AB}{8}$

$\Rightarrow AB=8\sin 57°=8\times 0.8387$

$\Rightarrow AB=6.7096cm$

Again,

$\Rightarrow \cos 57°=\frac{BC}{AC}=\frac{BC}{8}$

$\Rightarrow BC=8\times \cos 57°=8\times 0.5446$

$\Rightarrow BC=4.3568cm$

Now, area $△ABC=\frac{1}{2}\times BC\times AB$

$=\frac{1}{2}\times 4.3568\times 6.7096$

$=14.6161{cm}^2$

Hence, the answer is $14.6161{cm}^2.$