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Elden [556K]
4 years ago
15

Khan academywill mark.​

Mathematics
1 answer:
german4 years ago
8 0

Given:

\theta_1 is located in IV Quadrant.

\sin (\theta_1)=-\dfrac{24}{25}

To find:

The value of \cos \theta_1.

Solution:

We have,

\sin (\theta_1)=-\dfrac{24}{25}

We know that,

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

So,

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

\left(-\dfrac{24}{25}\right)^2+\cos^2(\theta_1)=1

\cos^2(\theta_1)=1-\dfrac{576}{625}

\cos^2(\theta_1)=\dfrac{625-576}{625}

Taking square root on both sides, we get

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

\cos (\theta_1)=\pm \dfrac{7}{25}

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

\cos (\theta_1)=\dfrac{7}{25}

Therefore, \cos (\theta_1)=\dfrac{7}{25}.

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