In triangle ABC, angle C is 90 °, AB = 5, BC = 3. Find the cosine of the outer angle at the vertex A.
| August 14, 2021 | education
Since triangle ABC is rectangular, then SinBAC = BC / AB = 3/5.
Determine the cosine of the angle BAC.
Cos2BAC = 1 – Sin2BAC = 1 – 9/25 = 25/25 – 9/25 = 16/25.
CosBAC = 4/5 = 0.8.
Outside angle BAD = (180 – BAC), then CosBAD = – CosBAC = -0.8.
Answer: The cosine of the outer corner is -0.8.
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