In a pre-rectangular triangle, the angle between the height and the meridian drawn from the apex of the right angle = 40 degrees, find the larger of the acute angles of this triangle.
In a right-angled triangle MBН, the angle НМВ = (180 – 90 – 40) = 50.
The HBM angle is adjacent to the AMB angle, then the AMB angle = (180 – 50) = 130.
The AMB triangle is isosceles by the median property, AM = BM.
Then the angle MAB = MBA = (180 – AMB) / 2 = (180 – 130) / 2 = 25.
Then the angle ACB = 90 – 25 = 65.
Answer: The larger acute angle is 65.
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