Given
To find the sine of each acute angle.
Now,
The sine of an angle in a right triangle is given by,
[tex]\sin \theta=\frac{opposite\text{ side}}{hypotnuse}[/tex]
In a right triangle ABC, angle A and angle C are the acute angles, since angle B is 90 degrees (right angle).
Then,
[tex]\begin{gathered} \sin A=\frac{BC}{AC} \\ =\frac{9}{41} \\ \sin C=\frac{AB}{AC} \\ =\frac{40}{41} \end{gathered}[/tex]
Hence, the sine of each acute angles in the triangle is 9/41 and 40/41.
