Given
To find the sine of each acute angle.
Now,
The sine of an angle in a right triangle is given by,
![\sin \theta=\frac{opposite\text{ side}}{hypotnuse}](https://img.qammunity.org/2023/formulas/mathematics/high-school/khm4xp7z8uehco3qq69p716k3v4e7kscrh.png)
In a right triangle ABC, angle A and angle C are the acute angles, since angle B is 90 degrees (right angle).
Then,
![\begin{gathered} \sin A=(BC)/(AC) \\ =(9)/(41) \\ \sin C=(AB)/(AC) \\ =(40)/(41) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/2kau2dcq1h5vvb75rtng87tst8tkgr3mpz.png)
Hence, the sine of each acute angles in the triangle is 9/41 and 40/41.