In a right-angled triangle MNK (angle K = 90 degrees) KN = 1dm, angle M = a. in what ratio does the height drawn to them divide the hypotenuse
Since AK is the height of the ABC triangle, then the AKM triangle is rectangular, and the angle KAM = 90, then the angle AKM = (90 – α) 0.
The angle HKM = 90, then the angle HKA = (90 – AKM) = (90 – 90 + α) = α0.
In a right-angled triangle AKH Sinα = AH / KH.
AH = Sinα * KH = Sinα * 1 = Sinα.
Cos α = AK / KH.
AK = Cosα * KH = Cosα * 1 = Cosα.
In a right-angled triangle AKM tgα = AK / AM.
AM = AK / tan α = Cosα / tanα = Cos2α / Sinα.
AH / AM = Sinα / (Cos2α / Sinα) = Sin2α / Cos2α = tg2α.
Answer: The ratio of the segments is tg2α.
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