In a right-angled triangle, the angle between the median and the bisector drawn from the apex of the right angle

In a right-angled triangle, the angle between the median and the bisector drawn from the apex of the right angle is 13 degrees. Find the larger of the two sharp corners of the triangle.

1. A, B, C – the vertices of the triangle. CК – median, СР – bisector. ∠С = 90 °. ∠KСР = 13 °.

2. The bisector CP divides ∠С into two equal angles:

∠ВСР = ∠АСР = 90 °: 2 = 45 °.

3. ∠АСК = ∠АСР – ∠КСР = 45 ° – 13 ° = 32 °.

4. The median CК is 1/2 hypotenuse AB. Therefore, AK = СK. That is, the AСK triangle

isosceles. The angles adjacent to the AC side are equal:

∠АВК = ∠САК (∠А) = 32 °.

5.∠В = 180 ° – ∠А – ∠С = 180 ° – 32 ° – 90 ° = 58 °.

Answer: ∠В = 58 ° – the larger acute angle of the triangle.