Question

If $\cos y^0=\sin (y^0+{20}^0)$, then find the least positive value of $y$.

Answer 1

$\cos y^0=\sin ({90}^0-y^0)$
$\therefore \sin ({90}^0-y^0)=\sin (y^0+{20}^0)$
or ${90}^0-y^0=y^0+{20}^0$
$\Rightarrow 2y^0={70}^{0}ory={35}^0$