In triangle AEC, angle E is 90 °, AC = 10, CE = 8. Find the sine of the outer corner at vertex C.

univerkov | February 14, 2021 | education

By the Pythagorean theorem, we determine the length of the leg AE.

AE ^ 2 = AC ^ 2 – CE ^ 2 = 100 – 64 = 36.

AE = 6 cm.

Then SinAEC = AE / AC = 6/10 = 3/5.

External angle ASD = (180 – ACE), then SinACD = Sin (180 – ACE) = SinACE = 3/5.

Second way.

Determine the cosine of the ACE angle.

CosACE = CE / AC = 8/10 = 4/5.

Then Sin2ACE = 1 – Cos2ACE = 1 – 16/25 = 9/25.

SinACE = 3/5.

Angle ACD = (180 – ACE), then SinACD = Sin (180 – ACE) = SinACE = 3/5.

Answer: The sine of the ACD angle is 3/5.

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