Question

Am I right? Please help me out

Answer 1

`\cos(\theta) = -(√(17))/(6)`

Explanation:

Given

`\tan(\theta) = -\sqrt{(19)/(17)}`

Required

Determine
`\cos(\theta)`

We have:

`\tan(\theta) = -\sqrt{(19)/(17)}`

Split

`\tan(\theta) = -(√(19))/(√(17))`

tan is calculated as:

`\tan(theta) = (opposite)/(adjacent)`

So:

`Opposite = -\sqrt{19`

`Adjacent = \sqrt{17`

And:

`Hypotenuse^2 = Opposite^2 + Adjacent^2` --- Pythagoras theorem

`Hypotenuse^2 = (-√(19))^2 + (√(17))^2`

`Hypotenuse^2 = 19 + 17`

`Hypotenuse^2 = 36`

Take square roots

`Hypotenuse = 6`

`\cos(\theta) = (Adjacent)/(Hypotenuse)`

`\cos(\theta) = (√(17))/(6)`

Since it is in the second quadrant, then:

`\cos(\theta) = -(√(17))/(6)`