If sinx = 0.6, and x is in quadrant ii, then what is the value of cosx?

To calculate for cos X given that sin X=0.6 and is in Quadrant we proceed as follows:
sin X=0.6
Thus implies that:
x=arcsin 0.6
x=36.97
given that X is in the Quadrant ii then the value of x will be:
180-36.97
=143.13
hence cos X will be:
cos X=cos 143.13
cos X=-0.8

 

cos(x)= -0.8

Step-by-step explanation:

If sinX = 0.6, and X is in Quadrant II,

sin(x)= 0.6 = 6/10

sin (x)= opposite side / hypotenuse

opposite side = 6 and hypotenuse = 10

use Pythagorean theorem

c^2 = a^2 + b^2

10^2 = 6^2 +b^2

100 = 36 + b^2

Subtract 36 from both sides

64= b^2

b= 8

Adjacent side = 8

Cos(x)= adjacent side / hypotenuse

cos(x)= 8/10= 0.8

Cos is negative in second quadrant

so cos(x)= -0.8

It would be 120 because 4% of 125 is 5, so you subtract 5 from 125 to get 120. I am pretty sure that's how you do it