In a right-angled triangle ABC leg AC = 65, and the height CH, lowered to the hypotenuse, is 13√21. Find sin∠ABC.

univerkov | September 26, 2021 | education

Triangles АСН and ВСН are similar in acute angle, angle СНН = АСН.

CosACH = CH / AC = 13 * √21 / 65 = √21 / 5.

Then Sin2ACH = 1 – Cos2ACH = 1 – 21/25 = 4/25

SinACH = SinCBH = SinABC = 2/5.

Answer: The sine of the angle ABC is 2/5.

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