theta=sin^-1 (5/13) ; (x,y)=(-3,1)
how do u find the coordinates of x' and y'??
To find the coordinates of x' and y', we need to apply some trigonometric principles.
Given the angle theta, which is equal to arcsin(5/13), we can find the sine and cosine of theta using the trigonometric identities.
To find sin(theta):
We know that sin(theta) = 5/13, so the opposite side is 5 and the hypotenuse is 13.
To find cos(theta):
We can use the Pythagorean theorem:
cos(theta) = sqrt(1 - sin^2(theta)) = sqrt(1 - (5/13)^2) = sqrt(1 - 25/169) = sqrt(144/169) = 12/13.
Now, let's consider the given point (x, y) = (-3, 1). This point lies in the third quadrant, where both x and y values are negative.
To find x' and y':
x' = x * cos(theta) - y * sin(theta)
y' = x * sin(theta) + y * cos(theta)
Plugging in the values:
x' = (-3) * (12/13) - (1) * (5/13) = -36/13 - 5/13 = -41/13
y' = (-3) * (5/13) + (1) * (12/13) = -15/13 + 12/13 = -3/13
Therefore, the coordinates of x' and y' are (-41/13, -3/13).