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sin A=3/5, A is at QII

Sagot :

sin = y/r so
y=3
r=5
x=?
To get the x we need to use the Pythagorean theorem
r^2=x^2 + y^2
transpose the y to the other side so we can only compute for the x
r^2 - y^2= x^2
then substitute the given above
5^2 - 3^2 = x^2
25 - 9 = x^2
√16=√x^2
±4=x
meaning the answer can be either +4 or -4 but since it lies on the QII where (-x,y) so the answer is -4