In a right-angled triangle ABC leg AC = 70, and the height CH lowered to the hypotenuse is 7√19, Find the sin of angle ABC
Triangles АСН and ВСН are similar in acute angle, angle СНН = АСН.
CosACH = CH / AC = 7 * √19 / 70 = √19 / 10.
Then Sin2ACH = 1 – Cos2ACH = 1 – 19/100 = 81/100.
SinACH = SinCBH = SinABC = 9/10.
Answer: The sine of the angle ABC is 9/10.
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