Question

If $\sin \theta =0.8$ and $\theta$ lies in $2^{nd}$quadrant. Find the value of $\cos \theta$.
Given, ${\sin }^2\theta +{\cos }^2\theta =1$

• A. $0.6$
• B. $-0.6$
• C. $-0.8$
• D. $0.8$

Answer 1

The correct option is B $-0.6$

We have, $\sin \theta =0.8$
And we have,
${\sin }^2\theta +{\cos }^2\theta =1\Rightarrow {\cos }^2\theta =1-{\sin }^2\theta \Rightarrow {\cos }^2\theta =1-{0.8}^2=0.36\Rightarrow {\cos }^2\theta =(0.6{)}^2\Rightarrow {\cos }^2\theta -(0.6{)}^2=0\Rightarrow (\cos \theta -0.6)(\cos \theta +0.6)=0\Rightarrow \cos \theta =0.6,-0.6$

Since $\theta$ lies in $2^{nd}$ quadrant and we know $\cos \theta$ is negative in $2^{nd}$ quadrant.
$\therefore \cos \theta =-0.6$