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

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

the answer is b.)

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