Question

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Answer 1

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