If the sides of a right –angled triangle are in AP. Then the sin  of the acute angles are

Let ABC be the right triangle, right angled at B. Let the sides of the triangle be $a–d,$ $a$ and $a+d,$ with $a>d>0.$ Clearly, $AC=a+d$ is the largest side. So, $a+d$ is the side opposite to the right angle?

By Pythagoras theorem, we have

  ${\left(a+d\right)}^2=a^2+{\left(a–d\right)}^2\Rightarrow 4ad=a^2\Rightarrow a=4d.$

Now, $\sin A=\frac{BC}{AC}=\frac{a}{a+d}=\frac{4d}{4d+d}=\frac{4}{5}$

And,  $\sin C=\frac{AB}{AC}=\frac{a–d}{a+d}=\frac{4d–d}{4d+d}=\frac{3}{5}$