In triangle ABC, angle C is 90 sina√ 7/4. Find sin B
| July 22, 2021 | education
In a right-angled triangle, the sine of one acute angle is equal to the cosine of another acute angle, then SinBAC = CosABC = √7 / 4.
Then: Sin2ABC = 1 – Cos2ABC = 1 – 7/16 = 9/16.
CosABC = 3/4.
Answer: The sine of angle B is 3/4.
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