SOLUTION: In a right triangle, the bisector of the right angle divides the hypotenuse in the ratio of 3 is to 5. Determine the measures of the acute angles of the triangle.

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Question 1008557: In a right triangle, the bisector of the right angle divides the hypotenuse in the ratio of 3 is to 5.
Determine the measures of the acute angles of the triangle.

Answer by ikleyn(52781) About Me  (Show Source):
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In a right triangle, the bisector of the right angle divides the hypotenuse in the ratio of 3 is to 5.
Determine the measures of the acute angles of the triangle.
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Let DELTAABC be our right-angled triangle with the right angle at the vertex A. 

Let AD be the bisector of the right angle, which intersects the hypotenuse at the point D.

The key to solving the problem is this Theorem of Planimetry:

    In a triangle, the angle bisector divides the side to which it is drawn, 
    in two segments proportional to the ratio of two other sides of a triangle.

The Theorem is valid for any triangle. For its proof see the lesson 
On what segments the angle bisector divides the side of a triangle in this site.


Particularly, for the given right-angled triangle DELTAABC the Theorem means that the ratio of the segment 
dimensions BD and CD is equal to the ratio of dimensions of the legs AB and AC: abs%28BD%29%2Fabs%28CD%29 = abs%28AB%29%2Fabs%28AC%29 = 3%2F5.

Thus the ratio of dimensions of the legs AB and AC is 3%2F5. 

One more time: abs%28AB%29%2Fabs%28AC%29 = 3%2F5.

It means that the tangent of the angle C of our triangle is 3%2F5:  tan(C) = 3%2F5.

In other words, the angle C of the triangle is atan%283%2F5%29:  C = atan%283%2F5%29

The problem is solved.