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

In a right-angled triangle, the angle between the height and the median drawn from the top of the corner to be taken is 14 degrees, find the smaller of the two acute angles of the triangle.

1. A, B, C – the vertices of the triangle. CE – height, СK – median. ∠С = 90 °. ∠EСK = 14 °.

2. ∠EX = 180 ° – ∠EСK – ∠СEC = 180 ° – 14 ° – 90 ° = 76 °.

3. ∠ВKС = 90 ° – 76 ° = 14 °.

4. The median CК, according to the properties of a right-angled triangle, is equal to 1/2 of the hypotenuse AB.

Therefore, ВK = СK. That is, the СВK triangle is isosceles. The angles adjacent to the BC side are equal:

∠СВК (∠В) = ∠ВСК = (180 ° – 14 °): 2 = 83 °.

5.∠А = 180 ° – ∠С – ∠В = 180 ° – 90 ° – 83 ° = 7 °.

Answer: ∠А = 7 ° – smaller acute angle.

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