In a right-angled triangle ABC with right angle C, the median CM is drawn sin ACM = √3 / 2, find sin B.
| April 2, 2021 | education
The median CM is drawn from the right angle ACB, then it is equal to half the length of the hypotenuse AB.
CM = AM = BM = AB / 2, and therefore the triangle AMC is isosceles.
Then SinACM = SinCAB = √3 / 2.
Determine the cosine of the angle CAB.
CosCAB ^ 2 = 1 – Sin ^ 2CAB = 1 – 3/4 = 1/4.
CosCAB = 1/2.
SinABC = AC / AB.
CosCAB = AC / AB.
SinABC = CosCAB = 1/2.
Answer: The sine of angle B is 1/2.
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