Question

If ${\sin }^2{19}^0+{\cos }^2\theta =1$ then. $\theta =$

• A. ${71}^0$
• B. ${19}^0$
• C. ${38}^0$
• D. ${171}^0$

Answer 1

Answer: (B)

${19}^0$

${\sin }^2{19}^{\circ }+{\cos }^2\theta =1$

$\Rightarrow {\cos }^2\theta =1-{\sin }^2{19}^{\circ }$

$\Rightarrow {\cos }^2\theta ={\cos }^2{19}^{\circ }$

$\Rightarrow (\cos \theta -\cos {19}^{\circ })(\cos \theta +\cos {19}^{\circ })=0$

$\Rightarrow (\cos \theta -\cos {19}^{\circ })=0$ or $(\cos \theta +\cos {19}^{\circ })=0$

$\Rightarrow \cos \theta =\cos {19}^{\circ }$ or $\cos \theta =-\cos {19}^{\circ }$

$\Rightarrow \cos \theta =\cos {19}^{\circ }$ or $\cos \theta =\cos ({180}^{\circ }-{19}^{\circ })$

$\therefore \theta ={19}^{\circ }$ or $\theta ={180}^{\circ }-{19}^{\circ }={161}^{\circ }$