Question

Khan academy

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Answer 1

Given:

`\theta_1` is located in IV Quadrant.

`\sin (\theta_1)=-(24)/(25)`

To find:

The value of
`\cos \theta_1`.

Solution:

We have,

`\sin (\theta_1)=-(24)/(25)`

We know that,

`\sin^2 \theta+\cos^2\theta=1`

So,

`\sin^2 (\theta_1)+\cos^2(\theta_1)=1`

`\left(-(24)/(25)\right)^2+\cos^2(\theta_1)=1`

`\cos^2(\theta_1)=1-(576)/(625)`

`\cos^2(\theta_1)=(625-576)/(625)`

Taking square root on both sides, we get

`\cos (\theta_1)=\pm \sqrt{(49)/(625)}`

`\cos (\theta_1)=\pm (7)/(25)`

`\theta_1` is located in IV Quadrant. In IV Quadrant the value of cos is positive. So,

`\cos (\theta_1)=(7)/(25)`

Therefore,
`\cos (\theta_1)=(7)/(25)`.