In a right-angled triangle, the angle between the height and the median, drawn from the top

In a right-angled triangle, the angle between the height and the median, drawn from the top of the right angle, is 26 °. Find the larger of the sharp corners of this triangle.

1. A, B, C – the vertices of the triangle. CE – height, СK – median. ∠С = 90 °. ∠EСK = 26 °.

2. ∠EКС = 180 ° – ∠EСK – ∠СEК = 180 ° – 26 ° – 90 ° = 64 °.

3.∠ВKС = 90 ° – 64 ° = 26 °. The median CК, according to the properties of a right-angled triangle, is

1/2 hypotenuse AB. Therefore, ВK = СK. That is, the СВK triangle is isosceles. Angles,

adjacent to the side of the aircraft are equal:

∠СВК (∠В) = ∠ВСК = (180 ° – 26 °): 2 = 77 °.

5.∠А = 180 ° – ∠С – ∠В = 180 ° – 90 ° – 77 ° = 13 °.

Answer: ∠В = 77 ° – a larger acute angle.