If cos@=15/19 and 270°<=360° , find sin@ and tan @?

To find sin@ and tan@, we can use the Pythagorean identity for trigonometric functions:

sin^2@ + cos^2@ = 1

Given that cos@ = 15/19, we can find sin@:

sin^2@ + (15/19)^2 = 1
sin^2@ + 225/361 = 1
sin^2@ = 1 - 225/361
sin^2@ = 136/361
sin@ = sqrt(136)/19

Now that we have sin@, we can find tan@:

tan@ = sin@ / cos@
tan@ = (sqrt(136)/19) / (15/19)
tan@ = sqrt(136) / 15

Therefore, sin@ = sqrt(136)/19 and tan@ = sqrt(136)/15.

Given
cos@=15/19 and 270°<=360°
1. Finding sin@
We know that sin^2 @+cos^2 @=1
=>sin^2 @=1-cos^2 @
=>sin^2 @=1-(15/19)^2
=>sin^2 @=1-225/361
=>sin^2 @=361/361-225/361
=>sin^2 @=136/361
=>sin@=√136/361
2. Finding tan@
We know that tan@=sin@/cos@
=>tan@=√136/361/15/19
=>tan@=(√136-19)/(361-15)
Answer
sin@=√136/361 , tan@=(√136-19)/(361-15)