Question

If $\sin ({30}^{\circ }-\theta )=\cos ({60}^{\circ }+\Phi )$ then

• A. $\Phi -\theta ={30}^{\circ }$
• B. $\Phi -\theta =0^{\circ }$
• C. $\Phi +\theta ={60}^{\circ }$
• D. $\Phi -\theta ={60}^{\circ }$

Answer 1

Answer: (B)

$\Phi -\theta =0^{\circ }$

$\sin ({30}^{\circ }-\theta )=\cos ({60}^{\circ }+\varphi )\Rightarrow \cos [{90}^{\circ }-({30}^{\circ }-\theta )]=\cos ({60}^{\circ }+\varphi )(\because \sin \theta =\cos ({90}^{\circ }-\theta ))\Rightarrow \cos ({60}^{\circ }+\theta )=\cos ({60}^{\circ }+\varphi )\Rightarrow {60}^{\circ }+\theta ={60}^{\circ }+\varphi \Rightarrow \theta -\varphi =0.$